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<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
<title>Siyuan's Blog</title>
<icon>https://www.gravatar.com/avatar/7f5c232e160bb21b4c009adb298fdcfa</icon>
<subtitle>你强归你强，我永不示弱！</subtitle>
<link href="/atom.xml" rel="self"/>
<link href="http://hydingsy.github.io/"/>
<updated>2019-03-09T12:55:59.253Z</updated>
<id>http://hydingsy.github.io/</id>
<author>
<name>Siyuan</name>
<email>294873684@qq.com</email>
</author>
<generator uri="http://hexo.io/">Hexo</generator>
<entry>
<title><strong><font size="20">「公告」博客搬迁</font></strong></title>
<link href="http://hydingsy.github.io/articles/Announcement/"/>
<id>http://hydingsy.github.io/articles/Announcement/</id>
<published>2019-03-06T16:00:00.000Z</published>
<updated>2019-03-09T12:55:59.253Z</updated>
<content type="html"><![CDATA[<center><font size="5">新博客地址：<a href="http://orzsiyuan.com" target="_blank" rel="noopener">orzsiyuan.com</a></font></center><p><strong>Siyuan’s Blog</strong>已经从 <a href="https://hydingsy.github.io">hydingsy.github.io</a> / <a href="http://old.orzsiyuan.com" target="_blank" rel="noopener">old.orzsiyuan.com</a> 搬迁到 <a href="http://orzsiyuan.com" target="_blank" rel="noopener">orzsiyuan.com</a>。本博客仅作为副本保留在此（这意味着今后可能删除），这里的文章将会陆续搬运到新博客中。</p><p><del>换过友链的大佬们可以考虑换个域名啦！新博客也继续资瓷换友链！QAQ</del></p><p>新的博客使用 <a href="http://typecho.org/" target="_blank" rel="noopener">Typecho</a> 搭建，欢迎大家访问！</p><div align="right">Siyuan</div><div align="right">2019.03.07</div>]]></content>
<summary type="html">
<center><font size="5">新博客地址：<a href="http://orzsiyuan.com" target="_blank" rel="noopener">orzsiyuan.com</a></font></center>
<p><strong>Siy
</summary>
</entry>
<entry>
<title>「Codeforces 402E」Strictly Positive Matrix</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-402E-Strictly-Positive-Matrix/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-402E-Strictly-Positive-Matrix/</id>
<published>2019-03-01T16:00:00.000Z</published>
<updated>2019-03-02T03:25:14.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/402/problem/E" target="_blank" rel="noopener">Codeforces 402E</a></p></blockquote><p>你有一个 $n\times n$ 的矩阵 $a$。它满足如下性质：</p><ul><li>对于任意的 $i,j$（$1\le i,j\le n$）都有 $a_{i,j}\ge 0$。</li><li>$\sum_{i=1}^n a_{i,i}=0$</li></ul><p>矩阵 $b$ 是严格正矩阵当且仅当对于任意的 $i,j$（$1\le i,j\le n$）都有 $b_{i,j}>0$。</p><p>求是否存在整数 $k\ge 1$ 使得 $a^k$ 是一个严格正矩阵。</p><p>数据范围：$2\le n\le 2\times 10^3$，$0\le a_{i,j}\le 50$，$\sum_{i=1}^n a_{i,i}=0$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>首先我们把这个矩阵转化为 $01$ 矩阵，它的 $k$ 次方代表着对它进行 $k$ 次迭代，得到的矩阵记为 $a’$。那么$a’_{i,j}=1$ 当且仅当<strong>邻接矩阵</strong> $a$ 形成的图中 $i$ 到 $j$ 有一条长度为 $k$ 的路径（并非是简单路径），也就是走 $k$ 次后图的连通性。</p><p>这样一来，我们把矩阵问题转化为图论问题了！</p><p>再观察这个问题的性质：由于对角线的和不为 $0$，那么存在至少一个点有<strong>自环</strong>。如果从 $i$ 到 $j$ 存在一条长度为 $k$ 且经过有自环的点，那么我们一定通过不断走自环可以得到长度为 $k+1,k+2,\dots$ 的路径。</p><p>这就意味着如果存在 $k$，那么一定<strong>所有点都可达</strong>。</p><p>直接用 $01$ 矩阵建立有向图，使用 $\text{Tarjan}$ 求出<strong>强连通分量</strong>的个数是否为 $1$ 即可。</p><p><strong>时间复杂度</strong>：$O(n^2)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><stack></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">define</span> fail puts(<span class="meta-string">"NO"</span>),exit(0)</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e3</span>+<span class="number">5</span>,M=N*N;</span><br><span class="line"><span class="keyword">int</span> n,idx,cnt,tot,lnk[N],ter[M],nxt[M],dfn[N],low[N];</span><br><span class="line"><span class="keyword">bool</span> vis[N];</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">stack</span><<span class="keyword">int</span>> st;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">add</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> v)</span> </span>{</span><br><span class="line"> ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">tarjan</span><span class="params">(<span class="keyword">int</span> u)</span> </span>{</span><br><span class="line"> dfn[u]=low[u]=++idx,st.push(u),vis[u]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(!dfn[v]) {</span><br><span class="line"> tarjan(v);</span><br><span class="line"> low[u]=<span class="built_in">std</span>::min(low[u],low[v]);</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> <span class="keyword">if</span>(vis[v]) low[u]=<span class="built_in">std</span>::min(low[u],dfn[v]);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(dfn[u]==low[u]) {</span><br><span class="line"> <span class="keyword">if</span>(++cnt><span class="number">1</span>) fail;</span><br><span class="line"> <span class="keyword">do</span> {</span><br><span class="line"> vis[u=st.top()]=<span class="number">0</span>,st.pop();</span><br><span class="line"> } <span class="keyword">while</span>(dfn[u]!=low[u]);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&n);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) {</span><br><span class="line"> <span class="keyword">int</span> x;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&x);</span><br><span class="line"> <span class="keyword">if</span>(x&&i!=j) add(i,j);</span><br><span class="line"> }</span><br><span class="line"> tarjan(<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(!dfn[i]) fail;</span><br><span class="line"> <span class="built_in">puts</span>(<span class="string">"YES"</span>);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/402/problem/E" target="_blank" rel="noopener">Codeforces 402E</a></p>
</blockquote>
<p>你有一个 $n\times n$ 的矩阵 $a$。它满足如下性质：</p>
<ul>
<li>对于任意的 $i,j$（$1\le i,j\le n$）都有 $a_{i,j}\ge 0$。</li>
<li>$\sum_{i=1}^n a_{i,i}=0$</li>
</ul>
<p>矩阵 $b$ 是严格正矩阵当且仅当对于任意的 $i,j$（$1\le i,j\le n$）都有 $b_{i,j}&gt;0$。</p>
<p>求是否存在整数 $k\ge 1$ 使得 $a^k$ 是一个严格正矩阵。</p>
<p>数据范围：$2\le n\le 2\times 10^3$，$0\le a_{i,j}\le 50$，$\sum_{i=1}^n a_{i,i}=0$</p>
</summary>
<category term="图论" scheme="http://hydingsy.github.io/tags/%E5%9B%BE%E8%AE%BA/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="邻接矩阵" scheme="http://hydingsy.github.io/tags/%E9%82%BB%E6%8E%A5%E7%9F%A9%E9%98%B5/"/>
<category term="Tarjan" scheme="http://hydingsy.github.io/tags/Tarjan/"/>
<category term="强连通分量" scheme="http://hydingsy.github.io/tags/%E5%BC%BA%E8%BF%9E%E9%80%9A%E5%88%86%E9%87%8F/"/>
</entry>
<entry>
<title>「Codeforces 1131D」Gourmet Choice</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-1131D-Gourmet-Choice/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-1131D-Gourmet-Choice/</id>
<published>2019-03-01T16:00:00.000Z</published>
<updated>2019-03-02T03:04:20.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/1131/problem/D" target="_blank" rel="noopener">Codeforces 1131D</a></p></blockquote><p>美食家 Apple 先生是一家美食杂志的主编。他会用一个正整数来评价每一道菜。</p><p>美食家在第一天品尝第 $n$ 道菜，第二天品尝了 $m$ 道菜。他制作了一张 $n\times m$ 的表格，记录了他对菜肴的评价。如果第一套中的第 $i$ 道菜比第二套中的第 $j$ 道菜好，那么 $a_{i,j}$ 等于 <code>></code>；如果要差，那么 $a_{i,j}$ 等于 <code><</code>。菜肴可能同样美味，那么 $a_{i,j}$ 等于 <code>=</code>。</p><p>现在 Apple 先生想让你帮他评价每道菜。由于他是非常严格的，他会对菜肴进行评估，以便使用的最大整数尽可能小。但是 Apple 先生也很公平，如果 $a_{i,j}$ 为 <code><</code>，那么给第一套中第 $i$ 道菜的评价一定小于第二套中第 $j$ 道菜。如果 $a_{i,j}$ 是 <code>></code> 那么应该要更大。如果 $a_{i,j}$ 为 <code>=</code>，那么这两个数字要相等。</p><p>帮助 Apple 先生评价这两套中的每一道菜，使之符合他的感受，并满足最大数字尽可能小。如果有解则输出 <code>Yes</code> 和评价的数字；否则输出 <code>No</code>。</p><p>数据范围：$1\le n,m\le 10^3$。</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>首先我们把 <code>=</code> 给处理掉，把所有相等的点缩在一起。如果同一个并查集中的点有 <code><</code> 或 <code>></code> 关系显然无解。</p><p>此时我们得到了一个有向图，直接对它<strong>拓扑排序</strong>，如果有环则无解，否则输出方案。</p><p><strong>时间复杂度</strong>：$O(nm)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdlib></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><queue></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">define</span> fail puts(<span class="meta-string">"No"</span>),exit(0)</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e3</span>+<span class="number">5</span>,M=<span class="number">2e6</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,cnt,tot,lnk[N],ter[M],nxt[M],fa[N],bel[N],deg[N],dis[N];</span><br><span class="line"><span class="keyword">bool</span> vis[N];</span><br><span class="line"><span class="keyword">char</span> s[N][N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">add</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> v)</span> </span>{</span><br><span class="line"> ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot,++deg[v];</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">find</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> fa[x]==x?x:fa[x]=find(fa[x]);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">init</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&m);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n+m;++i) fa[i]=i;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%s"</span>,s[i]+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) <span class="keyword">if</span>(s[i][j]==<span class="string">'='</span>) fa[find(i)]=find(n+j);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) {</span><br><span class="line"> <span class="keyword">if</span>(s[i][j]!=<span class="string">'='</span>&&find(i)==find(n+j)) fail;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n+m;++i) <span class="keyword">if</span>(find(i)==i) bel[i]=++cnt;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n+m;++i) bel[i]=bel[find(i)];</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">build</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) {</span><br><span class="line"> <span class="keyword">if</span>(s[i][j]==<span class="string">'<'</span>) add(bel[i],bel[j+n]);</span><br><span class="line"> <span class="keyword">if</span>(s[i][j]==<span class="string">'>'</span>) add(bel[j+n],bel[i]);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">queue</span><<span class="keyword">int</span>> q;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=cnt;++i) <span class="keyword">if</span>(!deg[i]) dis[i]=<span class="number">1</span>,q.push(i);</span><br><span class="line"> <span class="keyword">while</span>(!q.empty()) {</span><br><span class="line"> <span class="keyword">int</span> u=q.front(); q.pop();</span><br><span class="line"> vis[u]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> dis[v]=<span class="built_in">std</span>::max(dis[v],dis[u]+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">if</span>(!--deg[v]) q.push(v);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=cnt;++i) <span class="keyword">if</span>(!vis[i]) fail;</span><br><span class="line"> <span class="built_in">puts</span>(<span class="string">"Yes"</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n+m;++i) {</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d%c"</span>,dis[bel[i]],<span class="string">" \n"</span>[i==n||i==n+m]);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> init();</span><br><span class="line"> build();</span><br><span class="line"> solve();</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/1131/problem/D" target="_blank" rel="noopener">Codeforces 1131D</a></p>
</blockquote>
<p>美食家 Apple 先生是一家美食杂志的主编。他会用一个正整数来评价每一道菜。</p>
<p>美食家在第一天品尝第 $n$ 道菜，第二天品尝了 $m$ 道菜。他制作了一张 $n\times m$ 的表格，记录了他对菜肴的评价。如果第一套中的第 $i$ 道菜比第二套中的第 $j$ 道菜好，那么 $a_{i,j}$ 等于 <code>&gt;</code>；如果要差，那么 $a_{i,j}$ 等于 <code>&lt;</code>。菜肴可能同样美味，那么 $a_{i,j}$ 等于 <code>=</code>。</p>
<p>现在 Apple 先生想让你帮他评价每道菜。由于他是非常严格的，他会对菜肴进行评估，以便使用的最大整数尽可能小。但是 Apple 先生也很公平，如果 $a_{i,j}$ 为 <code>&lt;</code>，那么给第一套中第 $i$ 道菜的评价一定小于第二套中第 $j$ 道菜。如果 $a_{i,j}$ 是 <code>&gt;</code> 那么应该要更大。如果 $a_{i,j}$ 为 <code>=</code>，那么这两个数字要相等。</p>
<p>帮助 Apple 先生评价这两套中的每一道菜，使之符合他的感受，并满足最大数字尽可能小。如果有解则输出 <code>Yes</code> 和评价的数字；否则输出 <code>No</code>。</p>
<p>数据范围：$1\le n,m\le 10^3$。</p>
</summary>
<category term="图论" scheme="http://hydingsy.github.io/tags/%E5%9B%BE%E8%AE%BA/"/>
<category term="拓扑排序" scheme="http://hydingsy.github.io/tags/%E6%8B%93%E6%89%91%E6%8E%92%E5%BA%8F/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="缩点" scheme="http://hydingsy.github.io/tags/%E7%BC%A9%E7%82%B9/"/>
</entry>
<entry>
<title>「Codeforces 788E」New Task</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-788E-New-Task/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-788E-New-Task/</id>
<published>2019-02-27T16:00:00.000Z</published>
<updated>2019-02-28T23:59:44.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/788/problem/E" target="_blank" rel="noopener">Codeforces 788E</a></p></blockquote><p>在第 228 届国际 Uzhlyandian 战略比赛中，各队均由 $5$ 名队员组成。</p><p>Masha 是新的国防部长，这位部长的主要职责是计算军队的效率。军队由 $n$ 名士兵组成，标号为 $1$ 到 $n$，第 $i$ 个士兵的技能值为 $a_i$。</p><p>这个队伍将由 $3$ 名队员和 $2$ 名助手组成，队员的技能值是相同的，助手的技能值不大于队员的技能值。对于 Masha 而言，助手应该分别站在队员的最左侧和最右侧。形式化地说，一个队伍有 $5$ 名士兵，下标为 $i,j,k,l,p$ 满足 $1\le i<j<k<l<p\le n$ 并且 $a_i\le a_j=a_k=a_l\ge a_p$。</p><p>军队的效率是 Masha 可以选择的不同队伍的数量。两个队伍是不同的当且仅当有至少一个人不同。</p><p>最初，所有的士兵都可以成为队员。由于某些原因，有时一些士兵不能成为队员或可以成为队员；但是所有士兵在任何时候都可以成为助手。Masha 有 $m$ 次操作，每次可以令第 $x$ 个士兵可以成为队员或不能。她需要你求出每次操作后可以选择的队伍数量，答案对 $10^9+7$ 取模。</p><p>数据范围：$1\le n,m\le 10^5$，$1\le a_i\le 10^9$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>由于每个士兵都可以成为助手，我们设 $pre_i$ 和 $suf_i$ 分别表示 $i$ 左边和右边比 $a_i$ 小的数的个数。可以<strong>离散化</strong>后用树状数组求出。</p><p>由于有动态修改，并且区间看起来是可以合并的，我们考虑用线段树维护。</p><p>由于第 $1,5$ 个人无关紧要的，我们<strong>只关心中间的 $3$ 个人</strong>。又因为每个权值都是独立的，因此<strong>对每种权值建立一棵线段树</strong>（当然要采用动态开点）。</p><p>每个节点维护 $6$ 个值（其中 $p$ 为根节点，$[l,r]$ 为根节点维护的区间，$ls,rs$ 为左右儿子）：</p><table class="table table-bordered table-text-center table-vertical-middle"><thead> <tr><th>变量</th><th>含义</th><th>单点初值</th><th>转移</th></tr></thead><tbody> <tr><td>$A$</td><td>选择第 $1$ 个人的方案数</td><td>$pre_l$</td><td>$A_p=A_{ls}+A_{rs}$</td></tr> <tr><td>$B$</td><td>选择第 $2$ 个人的方案数</td><td>$1$</td><td>$B_p=B_{ls}+B_{rs}$</td></tr> <tr><td>$C$</td><td>选择第 $3$ 个人的方案数</td><td>$suf_r$</td><td>$C_p=C_{ls}+C_{rs}$</td></tr> <tr><td>$AB$</td><td>选择第 $1,2$ 个人的方案数</td><td>$0$</td><td>$AB_p=AB_{ls}+AB_{rs}+A_{ls}\cdot B_{rs}$</td></tr> <tr><td>$BC$</td><td>选择第 $2,3$ 个人的方案数</td><td>$0$</td><td>$BC_p=BC_{ls}+BC_{rs}+B_{ls}\cdot C_{rs}$</td></tr> <tr><td>$ABC$</td><td>选择第 $1,2,3$ 个人的方案数</td><td>$0$</td><td>$ABC_p=ABC_{ls}+ABC_{rs}+AB_{ls}\cdot C_{rs}+A_{ls}\cdot BC_{rs}$</td></tr></tbody></table><p>具体实现见代码。</p><p><strong>时间复杂度</strong>：$O((n+m)\log n)$（注意空间复杂度和主席树类似）</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">1e5</span>+<span class="number">5</span>,M=N*<span class="number">20</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> mod=<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,tot,a[N],b[N],pre[N],suf[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">upd</span><span class="params">(<span class="keyword">int</span> &x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> (x+=y)>=mod&&(x-=mod);</span><br><span class="line">}</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Bit</span> {</span></span><br><span class="line"> <span class="keyword">int</span> b[N];</span><br><span class="line"> <span class="function"><span class="keyword">void</span> <span class="title">init</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">memset</span>(b,<span class="number">0</span>,<span class="keyword">sizeof</span>(b));</span><br><span class="line"> }</span><br><span class="line"> <span class="function"><span class="keyword">void</span> <span class="title">modify</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> v)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(;x<=n;x+=x&-x) b[x]+=v;</span><br><span class="line"> }</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">query</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(;x;x^=x&-x) ans+=b[x];</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line"> }</span><br><span class="line">} bit;</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Segment</span> {</span></span><br><span class="line"> <span class="keyword">int</span> rt[N],ls[M],rs[M],A[M],B[M],C[M],AB[M],BC[M],ABC[M];</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">plus</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> c=x+y;</span><br><span class="line"> <span class="keyword">return</span> c>=mod?c-mod:c;</span><br><span class="line"> }</span><br><span class="line"> <span class="function"><span class="keyword">int</span> <span class="title">mul</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> <span class="number">1L</span>L*x*y%mod;</span><br><span class="line"> }</span><br><span class="line"> <span class="function"><span class="keyword">void</span> <span class="title">pushup</span><span class="params">(<span class="keyword">int</span> p)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> l=ls[p],r=rs[p];</span><br><span class="line"> A[p]=plus(A[l],A[r]);</span><br><span class="line"> B[p]=plus(B[l],B[r]);</span><br><span class="line"> C[p]=plus(C[l],C[r]);</span><br><span class="line"> AB[p]=plus(plus(AB[l],AB[r]),mul(A[l],B[r]));</span><br><span class="line"> BC[p]=plus(plus(BC[l],BC[r]),mul(B[l],C[r]));</span><br><span class="line"> ABC[p]=plus(plus(ABC[l],ABC[r]),plus(mul(A[l],BC[r]),mul(AB[l],C[r])));</span><br><span class="line"> }</span><br><span class="line"> <span class="function"><span class="keyword">void</span> <span class="title">modify</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> &p,<span class="keyword">int</span> l,<span class="keyword">int</span> r,<span class="keyword">int</span> v)</span> </span>{</span><br><span class="line"> <span class="keyword">if</span>(!p) p=++tot;</span><br><span class="line"> <span class="keyword">if</span>(l==r) {</span><br><span class="line"> A[p]=v*pre[x];</span><br><span class="line"> B[p]=v;</span><br><span class="line"> C[p]=v*suf[x];</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">int</span> mid=(l+r)>><span class="number">1</span>;</span><br><span class="line"> <span class="keyword">if</span>(x<=mid) modify(x,ls[p],l,mid,v);</span><br><span class="line"> <span class="keyword">else</span> modify(x,rs[p],mid+<span class="number">1</span>,r,v);</span><br><span class="line"> pushup(p);</span><br><span class="line"> }</span><br><span class="line">} seg;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">init</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">std</span>::sort(b+<span class="number">1</span>,b+n+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">int</span> sz=<span class="built_in">std</span>::unique(b+<span class="number">1</span>,b+n+<span class="number">1</span>)-(b+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) a[i]=<span class="built_in">std</span>::lower_bound(b+<span class="number">1</span>,b+sz+<span class="number">1</span>,a[i])-b;</span><br><span class="line"> bit.init();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) pre[i]=bit.query(a[i]),bit.modify(a[i],<span class="number">1</span>);</span><br><span class="line"> bit.init();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=n;i>=<span class="number">1</span>;--i) suf[i]=bit.query(a[i]),bit.modify(a[i],<span class="number">1</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&n);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&a[i]),b[i]=a[i];</span><br><span class="line"> init();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) seg.modify(i,seg.rt[a[i]],<span class="number">1</span>,n,<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) upd(ans,seg.ABC[seg.rt[i]]);</span><br><span class="line"> <span class="keyword">for</span>(<span class="built_in">scanf</span>(<span class="string">"%d"</span>,&m);m--;) {</span><br><span class="line"> <span class="keyword">int</span> opt,x;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&opt,&x);</span><br><span class="line"> <span class="keyword">int</span> &rt=seg.rt[a[x]];</span><br><span class="line"> upd(ans,mod-seg.ABC[rt]);</span><br><span class="line"> seg.modify(x,rt,<span class="number">1</span>,n,opt<span class="number">-1</span>);</span><br><span class="line"> upd(ans,seg.ABC[rt]);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/788/problem/E" target="_blank" rel="noopener">Codeforces 788E</a></p>
</blockquote>
<p>在第 228 届国际 Uzhlyandian 战略比赛中，各队均由 $5$ 名队员组成。</p>
<p>Masha 是新的国防部长，这位部长的主要职责是计算军队的效率。军队由 $n$ 名士兵组成，标号为 $1$ 到 $n$，第 $i$ 个士兵的技能值为 $a_i$。</p>
<p>这个队伍将由 $3$ 名队员和 $2$ 名助手组成，队员的技能值是相同的，助手的技能值不大于队员的技能值。对于 Masha 而言，助手应该分别站在队员的最左侧和最右侧。形式化地说，一个队伍有 $5$ 名士兵，下标为 $i,j,k,l,p$ 满足 $1\le i&lt;j&lt;k&lt;l&lt;p\le n$ 并且 $a_i\le a_j=a_k=a_l\ge a_p$。</p>
<p>军队的效率是 Masha 可以选择的不同队伍的数量。两个队伍是不同的当且仅当有至少一个人不同。</p>
<p>最初，所有的士兵都可以成为队员。由于某些原因，有时一些士兵不能成为队员或可以成为队员；但是所有士兵在任何时候都可以成为助手。Masha 有 $m$ 次操作，每次可以令第 $x$ 个士兵可以成为队员或不能。她需要你求出每次操作后可以选择的队伍数量，答案对 $10^9+7$ 取模。</p>
<p>数据范围：$1\le n,m\le 10^5$，$1\le a_i\le 10^9$</p>
</summary>
<category term="数据结构" scheme="http://hydingsy.github.io/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84/"/>
<category term="线段树" scheme="http://hydingsy.github.io/tags/%E7%BA%BF%E6%AE%B5%E6%A0%91/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="树状数组" scheme="http://hydingsy.github.io/tags/%E6%A0%91%E7%8A%B6%E6%95%B0%E7%BB%84/"/>
<category term="计数" scheme="http://hydingsy.github.io/tags/%E8%AE%A1%E6%95%B0/"/>
</entry>
<entry>
<title>「UOJ 454」打雪仗</title>
<link href="http://hydingsy.github.io/articles/problem-UOJ-454-Snowball-Fight/"/>
<id>http://hydingsy.github.io/articles/problem-UOJ-454-Snowball-Fight/</id>
<published>2019-02-26T16:00:00.000Z</published>
<updated>2019-03-02T03:54:12.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="http://uoj.ac/problem/454" target="_blank" rel="noopener">UOJ 454</a></p></blockquote><p>这是一道通信题。</p><p>小 A 有一个长度为 $2n$ 的 $01$ 字符串 $S$，小 B 有 $\{1,\dots,2n\}$ 这些下标中的 $n$ 个：$p_1,\dots,p_n$。小 A 和小 B 可以相互之间可以发送字符（但只能发送 <code>0</code> 或者 <code>1</code> 两种）。请为小 A 和小 B 设计一种通信方式，使得小 B 最终能知道这 $n$ 个下标所对应的字符 $S[p_1],S[p_2],\dots,S[p_n]$。两人中发送字符数较多的那一个不应发送超过 $m$ 个字符。</p><p>数据范围：$n=1000$，$m=1350$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>我们把这 $2n$ 个下标分为 $k$ 块，定义第 $i$ 个块中小 B 需要的下标有 $s_i$ 个。由于小 B 知道的 $n$ 个下标是总数的一半，那么至少有一个块中 $s_i$ 不小于块的长度的一半。我们考虑这样一个策略：</p><ul><li><p>把这 $k$ 个块中的其中 $c$ 个块让小 A 暴力发送给小 B。其余块中的下标让小 B 告诉小 A 是否要发送。</p></li><li><p>为了最大化暴力发送的效率，我们要选择 $s_i$ 最大的那 $c$ 个块。考虑最坏情况：$s_i$ 均为块长度的一半。</p></li><li><p>通过简单计算我们可以得到：</p><p>小 A 的发送次数：$n+\frac{c}{k}\cdot n$</p><p>小 B 的发送次数：$2n-\frac{c}{k}\cdot 2n$</p></li><li><p>计算不等式得到：$0.325\le \frac{c}{k}\le 0.35$。</p></li><li><p>这样一来我们取 $\frac{c}{k}=\frac{1}{3}=0.333\dots$ 就行了。</p></li></ul><p><strong>时间复杂度</strong>：$O(n)$</p><p><strong>发送字符数</strong>：不超过 $\frac{4}{3}n$ 个字符。</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><h3 id="Alice"><a href="#Alice" class="headerlink" title="Alice"></a>Alice</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e3</span>+<span class="number">5</span>;</span><br><span class="line">FILE *in=fopen(<span class="string">"alice.in"</span>,<span class="string">"r"</span>);</span><br><span class="line"><span class="keyword">int</span> n;</span><br><span class="line"><span class="keyword">char</span> s[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">send</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="built_in">putchar</span>(x+<span class="string">'0'</span>),fflush(<span class="built_in">stdout</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">get</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> getchar()^<span class="number">48</span>;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">(<span class="keyword">int</span> l,<span class="keyword">int</span> r)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=l;i<=r;++i) send(s[i]-<span class="string">'0'</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(i<l||i>r) {</span><br><span class="line"> <span class="keyword">if</span>(get()) send(s[i]-<span class="string">'0'</span>);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">fscanf</span>(in,<span class="string">"%d%*d%s"</span>,&n,s+<span class="number">1</span>),n<<=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">int</span> l1=<span class="number">1</span>,r1=n/<span class="number">3</span>,l2=r1+<span class="number">1</span>,r2=n/<span class="number">3</span>*<span class="number">2</span>,l3=r2+<span class="number">1</span>,r3=n;</span><br><span class="line"> <span class="keyword">int</span> opt=<span class="number">2</span>*get()+get();</span><br><span class="line"> <span class="keyword">if</span>(opt==<span class="number">1</span>) solve(l1,r1);</span><br><span class="line"> <span class="keyword">if</span>(opt==<span class="number">2</span>) solve(l2,r2);</span><br><span class="line"> <span class="keyword">if</span>(opt==<span class="number">3</span>) solve(l3,r3);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h3 id="Bob"><a href="#Bob" class="headerlink" title="Bob"></a>Bob</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line">{</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e3</span>+<span class="number">5</span>;</span><br><span class="line">FILE *in=fopen(<span class="string">"bob.in"</span>,<span class="string">"r"</span>),*out=fopen(<span class="string">"bob.out"</span>,<span class="string">"w"</span>);</span><br><span class="line"><span class="keyword">int</span> n,p[N],vis[N],ans[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">send</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="built_in">putchar</span>(x+<span class="string">'0'</span>),fflush(<span class="built_in">stdout</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">get</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> getchar()^<span class="number">48</span>;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">calc</span><span class="params">(<span class="keyword">int</span> l,<span class="keyword">int</span> r)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=l;i<=r;++i) ans+=vis[i];</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">(<span class="keyword">int</span> l,<span class="keyword">int</span> r,<span class="keyword">int</span> o1,<span class="keyword">int</span> o2)</span> </span>{</span><br><span class="line"> send(o1),send(o2);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=l;i<=r;++i) ans[i]=get();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(i<l||i>r) {</span><br><span class="line"> send(vis[i]);</span><br><span class="line"> <span class="keyword">if</span>(vis[i]) ans[i]=get();</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">fscanf</span>(in,<span class="string">"%d%*d"</span>,&n);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="built_in">fscanf</span>(in,<span class="string">"%d"</span>,&p[i]);</span><br><span class="line"> vis[p[i]]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> n<<=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">int</span> l1=<span class="number">1</span>,r1=n/<span class="number">3</span>,l2=r1+<span class="number">1</span>,r2=n/<span class="number">3</span>*<span class="number">2</span>,l3=r2+<span class="number">1</span>,r3=n;</span><br><span class="line"> <span class="keyword">int</span> c1=calc(l1,r1),c2=calc(l2,r2),c3=calc(l3,r3);</span><br><span class="line"> <span class="keyword">int</span> mx=<span class="built_in">std</span>::max(c1,<span class="built_in">std</span>::max(c2,c3));</span><br><span class="line"> mx==c1?solve(l1,r1,<span class="number">0</span>,<span class="number">1</span>):mx==c2?solve(l2,r2,<span class="number">1</span>,<span class="number">0</span>):solve(l3,r3,<span class="number">1</span>,<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(vis[i]) <span class="built_in">fprintf</span>(out,<span class="string">"%d"</span>,ans[i]);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="http://uoj.ac/problem/454" target="_blank" rel="noopener">UOJ 454</a></p>
</blockquote>
<p>这是一道通信题。</p>
<p>小 A 有一个长度为 $2n$ 的 $01$ 字符串 $S$，小 B 有 $\{1,\dots,2n\}$ 这些下标中的 $n$ 个：$p_1,\dots,p_n$。小 A 和小 B 可以相互之间可以发送字符（但只能发送 <code>0</code> 或者 <code>1</code> 两种）。请为小 A 和小 B 设计一种通信方式，使得小 B 最终能知道这 $n$ 个下标所对应的字符 $S[p_1],S[p_2],\dots,S[p_n]$。两人中发送字符数较多的那一个不应发送超过 $m$ 个字符。</p>
<p>数据范围：$n=1000$，$m=1350$</p>
</summary>
<category term="UOJ" scheme="http://hydingsy.github.io/tags/UOJ/"/>
<category term="贪心" scheme="http://hydingsy.github.io/tags/%E8%B4%AA%E5%BF%83/"/>
<category term="构造" scheme="http://hydingsy.github.io/tags/%E6%9E%84%E9%80%A0/"/>
<category term="通信题" scheme="http://hydingsy.github.io/tags/%E9%80%9A%E4%BF%A1%E9%A2%98/"/>
</entry>
<entry>
<title>「APIO 2016」Gap</title>
<link href="http://hydingsy.github.io/articles/problem-APIO-2016-Gap/"/>
<id>http://hydingsy.github.io/articles/problem-APIO-2016-Gap/</id>
<published>2019-02-26T16:00:00.000Z</published>
<updated>2019-02-28T03:06:46.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="">UOJ 206</a></p></blockquote><p>这是一道交互题。</p><p>有 $n$ 个严格递增的非负整数 $a_1,a_2,\dots,a_n$（$0\le a_1<a_2<\dots<a_n\le 10^{18}$）。你需要找出 $a_{i+1}−a_i$（$1\le i\le n-1$）里的最大的值。</p><p>你的程序不能直接读入这个整数序列，但是你可以通过给定的函数来查询该序列的信息。关于查询函数的细节，请根据你所使用的语言，参考下面的实现细节部分。</p><p>你需要实现一个函数，该函数返回 $a_{i+1}−a_i$（$1\le i\le n-1$）中的最大值。</p><h3 id="实现细节（C-C-）"><a href="#实现细节（C-C-）" class="headerlink" title="实现细节（C/C++）"></a>实现细节（C/C++）</h3><p>你需要包含头文件 <code>gap.h</code>。</p><p>你需要实现一个函数 $\text{findGap}(T,N)$，该函数接受下面的参数，并返回一个 $\text{long long}$ 类型的整数：</p><ul><li>$T$：子任务的编号（$1$ 或者 $2$）</li><li>$N$：序列的长度</li></ul><p>你的函数 $\text{findGap}$ 可以调用系统提供的查询函数 $\text{MinMax}(s,t,\&mn,\&mx)$，该函数的前两个参数 $s$ 和 $t$ 是 $\text{long long}$ 类型的整数，后两个参数 $\&mn$ 和 $\&mx$ 是 $\text{long long}$ 类型的整数的指针（$mn$ 和 $mx$ 是 $\text{long long}$ 类型的整数）。当 $\text{MinMax}(s,t,\&mn,\&mx)$ 返回时，变量 $mn$ 将会存储满足 $a_i\in [s,t]$ 中 $a_i$ 的最小值，变量 $mx$ 将会存储满足 $a_i\in[s,t]$ 中 $a_i$ 的最大值。如果区间 $[s,t]$ 中没有序列中的数，则 $mn$ 和 $mx$ 都将存储 $-1$。在查询时需要满足 $s\le t$，否则程序将会终止，该测试点计为 $0$ 分。</p><h3 id="限制与约定"><a href="#限制与约定" class="headerlink" title="限制与约定"></a>限制与约定</h3><p>对于所有的测试点，有 $2\le n\le 10^5$。</p><p>每一个测试点开始测试之前，$M$ 都将被初始化为 $0$。</p><p>子任务 $1$（$30$ 分）：每一次调用 $\text{MinMax}$ 都将使 $M$ 加 $1$。为了获得所有分数，需要满足对于该子任务下的所有测试点，都有 $M\le \frac{n+1}{2}$。</p><p>子任务 $2$（$70$ 分）：定义 $k$ 为调用 $\text{MinMax}$ 时，区间 $[s,t]$ 中的序列中数的数量。每次调用 $\text{MinMax}$，将使 $M$ 加上 $k+1$。对于每一个测试点，如果 $M\le 3n$，你将得到 $70$ 分，否则将得到 $\frac{60}{\sqrt{\frac{M}{n}+1}-1}$ 分。你的该子任务的得分是其下所有测试点中的最低分。</p><h3 id="下载"><a href="#下载" class="headerlink" title="下载"></a>下载</h3><p><a href="http://uoj.ac/download.php?type=problem&id=206" target="_blank" rel="noopener">样例及测评库下载</a></p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><h3 id="子任务-1"><a href="#子任务-1" class="headerlink" title="子任务 1"></a>子任务 1</h3><p>由于有 $0\le a_1<a_2<\dots<a_n\le 10^{18}$ 这个限制，我们可以<strong>从两头向中间</strong>询问，第一次询问 $[0,10^{18}]$ 得到 $a_1,a_n$ 的值，接下来询问 $[a_1+1,a_n-1]$ 得到 $a_2,a_{n-1}$ 的值……以此类推，我们可以发现询问次数恰好是 $\frac{n+1}{2}$，可以通过本子任务。</p><h3 id="子任务-2"><a href="#子任务-2" class="headerlink" title="子任务 2"></a>子任务 2</h3><p>我们先确定<strong>答案的下界</strong>：设 $s,t$ 为询问 $[0,10^{18}]$ 得到的结果，那么通过<strong>鸽巢原理</strong>可以得到答案的下界为 $\left\lfloor\frac{t-s}{n-1}\right\rfloor$，记为 $p$（两个数之间的最大间隔最少为 $p$）。这样一来，我们就<strong>只关心那些长度为 $p+1$ 的区间</strong>了。</p><p>设当前询问的区间为 $[l,r]$（初始值 $l=s+1$，且显然有 $r=l+p$），得到的结果为 $mn,mx$（当 $mn=-1,mx=-1$ 时忽略这个区间）；那么我们用 $mn-\text{上一个出现的数字}$ 来更新答案，并将 $\text{上一个出现的数字}$ 设为 $mx$。接下来询问区间 $[l’,r’]$（其中 $l’=r+1,r’=l’+p$），不断更新答案直到 $l>t$ 为止。</p><p>最后证明一下 $M\le 3n$。第一次询问得到 $[s,t]$ 后 $k=n$，接下来不超过 $n-1$ 次询问，总共包含不超过 $n$ 个点，故 $M\le (n+1)+(n-1)+n\le 3n$，符合条件。</p><p> <strong>时间复杂度</strong>：$O(n)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string">"gap.h"</span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">1e5</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> a[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">findGap</span><span class="params">(<span class="keyword">int</span> T,<span class="keyword">int</span> n)</span> </span>{</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">if</span>(T==<span class="number">1</span>) {</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> s=<span class="number">0</span>,t=<span class="number">1e18</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> l=<span class="number">1</span>,r=n;l<=r;++l,--r) MinMax(s,t,&a[l],&a[r]),s=a[l]+<span class="number">1</span>,t=a[r]<span class="number">-1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i<=n;++i) ans=<span class="built_in">std</span>::max(ans,a[i]-a[i<span class="number">-1</span>]);</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> s,t;</span><br><span class="line"> MinMax(<span class="number">0</span>,<span class="number">1e18</span>,&s,&t);</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> p=(t-s)/(n<span class="number">-1</span>),pre=s;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">long</span> <span class="keyword">long</span> i=s+<span class="number">1</span>;i<=t;i+=p+<span class="number">1</span>) {</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> mn,mx;</span><br><span class="line"> MinMax(i,i+p,&mn,&mx);</span><br><span class="line"> <span class="keyword">if</span>(~mn) ans=<span class="built_in">std</span>::max(ans,mn-pre),pre=mx;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="">UOJ 206</a></p>
</blockquote>
<p>这是一道交互题。</p>
<p>有 $n$ 个严格递增的非负整数 $a_1,a_2,\dots,a_n$（$0\le a_1&lt;a_2&lt;\dots&lt;a_n\le 10^{18}$）。你需要找出 $a_{i+1}−a_i$（$1\le i\le n-1$）里的最大的值。</p>
<p>你的程序不能直接读入这个整数序列，但是你可以通过给定的函数来查询该序列的信息。关于查询函数的细节，请根据你所使用的语言，参考下面的实现细节部分。</p>
<p>你需要实现一个函数，该函数返回 $a_{i+1}−a_i$（$1\le i\le n-1$）中的最大值。</p>
<h3 id="实现细节（C-C-）"><a href="#实现细节（C-C-）" class="headerlink" title="实现细节（C/C++）"></a>实现细节（C/C++）</h3><p>你需要包含头文件 <code>gap.h</code>。</p>
<p>你需要实现一个函数 $\text{findGap}(T,N)$，该函数接受下面的参数，并返回一个 $\text{long long}$ 类型的整数：</p>
<ul>
<li>$T$：子任务的编号（$1$ 或者 $2$）</li>
<li>$N$：序列的长度</li>
</ul>
<p>你的函数 $\text{findGap}$ 可以调用系统提供的查询函数 $\text{MinMax}(s,t,\&amp;mn,\&amp;mx)$，该函数的前两个参数 $s$ 和 $t$ 是 $\text{long long}$ 类型的整数，后两个参数 $\&amp;mn$ 和 $\&amp;mx$ 是 $\text{long long}$ 类型的整数的指针（$mn$ 和 $mx$ 是 $\text{long long}$ 类型的整数）。当 $\text{MinMax}(s,t,\&amp;mn,\&amp;mx)$ 返回时，变量 $mn$ 将会存储满足 $a_i\in [s,t]$ 中 $a_i$ 的最小值，变量 $mx$ 将会存储满足 $a_i\in[s,t]$ 中 $a_i$ 的最大值。如果区间 $[s,t]$ 中没有序列中的数，则 $mn$ 和 $mx$ 都将存储 $-1$。在查询时需要满足 $s\le t$，否则程序将会终止，该测试点计为 $0$ 分。</p>
<h3 id="限制与约定"><a href="#限制与约定" class="headerlink" title="限制与约定"></a>限制与约定</h3><p>对于所有的测试点，有 $2\le n\le 10^5$。</p>
<p>每一个测试点开始测试之前，$M$ 都将被初始化为 $0$。</p>
<p>子任务 $1$（$30$ 分）：每一次调用 $\text{MinMax}$ 都将使 $M$ 加 $1$。为了获得所有分数，需要满足对于该子任务下的所有测试点，都有 $M\le \frac{n+1}{2}$。</p>
<p>子任务 $2$（$70$ 分）：定义 $k$ 为调用 $\text{MinMax}$ 时，区间 $[s,t]$ 中的序列中数的数量。每次调用 $\text{MinMax}$，将使 $M$ 加上 $k+1$。对于每一个测试点，如果 $M\le 3n$，你将得到 $70$ 分，否则将得到 $\frac{60}{\sqrt{\frac{M}{n}+1}-1}$ 分。你的该子任务的得分是其下所有测试点中的最低分。</p>
<h3 id="下载"><a href="#下载" class="headerlink" title="下载"></a>下载</h3><p><a href="http://uoj.ac/download.php?type=problem&amp;id=206" target="_blank" rel="noopener">样例及测评库下载</a></p>
</summary>
<category term="APIO" scheme="http://hydingsy.github.io/tags/APIO/"/>
<category term="UOJ" scheme="http://hydingsy.github.io/tags/UOJ/"/>
<category term="交互题" scheme="http://hydingsy.github.io/tags/%E4%BA%A4%E4%BA%92%E9%A2%98/"/>
<category term="鸽巢原理" scheme="http://hydingsy.github.io/tags/%E9%B8%BD%E5%B7%A2%E5%8E%9F%E7%90%86/"/>
</entry>
<entry>
<title>「WC 2015」未来程序</title>
<link href="http://hydingsy.github.io/articles/problem-WC-2015-Future-Program/"/>
<id>http://hydingsy.github.io/articles/problem-WC-2015-Future-Program/</id>
<published>2019-02-25T16:00:00.000Z</published>
<updated>2019-03-03T02:42:42.139Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="http://uoj.ac/problem/73" target="_blank" rel="noopener">UOJ 73</a></p></blockquote><p>本题是一道提交答案题，一共有 $10$ 个测试点。</p><p>对于每个测试点，你会得到一段程序的源代码和这段程序的输入。你要运行这个程序，并保存这个程序的输出。</p><p>遗憾的是这些程序效率都极其低下，无法在比赛的 $5$ 小时内得到输出。</p><p>你需要帮助 B 君得到这些程序的输出。</p><p><a href="http://uoj.ac/download.php?type=problem&id=73" target="_blank" rel="noopener">输入数据下载</a></p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><h3 id="Program-1"><a href="#Program-1" class="headerlink" title="Program 1"></a>Program 1</h3><h4 id="分析"><a href="#分析" class="headerlink" title="分析"></a>分析</h4><p>给 $d$ 加上 $b$ 个 $a$，答案对 $c$ 取模（注意模数很大）。</p><h4 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> a,b,mod;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">mul</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> x,<span class="keyword">long</span> <span class="keyword">long</span> p)</span> </span>{</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(;p;p>>=<span class="number">1</span>,x=(x+x)%mod) <span class="keyword">if</span>(p&<span class="number">1</span>) ans=(ans+x)%mod;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> freopen(<span class="string">"program1.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program1.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="keyword">while</span>(~<span class="built_in">scanf</span>(<span class="string">"%lld%lld%lld"</span>,&a,&b,&mod)) {</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%lld\n"</span>,mul(a,b,mod));</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案"><a href="#答案" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">11239440904485</span></span><br><span class="line"><span class="number">7551029211890</span></span><br><span class="line"><span class="number">20677492996370</span></span><br><span class="line"><span class="number">592966462292420</span></span><br><span class="line"><span class="number">69231182718627</span></span><br><span class="line"><span class="number">479525534330380</span></span><br><span class="line"><span class="number">544015996901435</span></span><br><span class="line"><span class="number">214227311823605</span></span><br><span class="line"><span class="number">73749675429767</span></span><br><span class="line"><span class="number">239498441843796</span></span><br></pre></td></tr></table></figure><h3 id="Program-2"><a href="#Program-2" class="headerlink" title="Program 2"></a>Program 2</h3><h4 id="分析-1"><a href="#分析-1" class="headerlink" title="分析"></a>分析</h4><p>可以找出 $a,b,c$ 的递推式子：</p><script type="math/tex; mode=display">a\leftarrow 1a+2b+1c\\b\leftarrow 1a+1b+0c\\c\leftarrow 1a+0b+0c\\</script><p>得到转移矩阵：</p><script type="math/tex; mode=display">\begin{bmatrix}1 & 2 & 1 \\1 & 1 & 0 \\1 & 0 & 0\end{bmatrix}</script><p>直接矩阵快速幂即可解决本题。</p><h4 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">4</span>;</span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> n,mod;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">upd</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> &x,<span class="keyword">long</span> <span class="keyword">long</span> y)</span> </span>{</span><br><span class="line"> (x+=y)>=mod&&(x-=mod);</span><br><span class="line">}</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Matrix</span> {</span></span><br><span class="line"> <span class="keyword">int</span> n;</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> A[N][N];</span><br><span class="line"> Matrix(<span class="keyword">int</span> _n=<span class="number">0</span>) {</span><br><span class="line"> n=_n,<span class="built_in">memset</span>(A,<span class="number">0</span>,<span class="keyword">sizeof</span>(A));</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">void</span> <span class="keyword">operator</span> ~ () {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) A[i][i]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> Matrix <span class="keyword">operator</span> * (<span class="keyword">const</span> Matrix &b)<span class="keyword">const</span> {</span><br><span class="line"> <span class="function">Matrix <span class="title">ans</span><span class="params">(n)</span></span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">1</span>;k<=n;++k) {</span><br><span class="line"> upd(ans.A[i][j],A[i][k]*b.A[k][j]%mod);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line"> }</span><br><span class="line"> Matrix <span class="keyword">operator</span> ^ (<span class="keyword">const</span> <span class="keyword">long</span> <span class="keyword">long</span> &b)<span class="keyword">const</span> {</span><br><span class="line"> Matrix ans(n),x=*this; ~ans;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">long</span> <span class="keyword">long</span> p=b;p;p>>=<span class="number">1</span>,x=x*x) <span class="keyword">if</span>(p&<span class="number">1</span>) ans=ans*x;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line"> }</span><br><span class="line">};</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> freopen(<span class="string">"program2.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program2.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="keyword">while</span>(~<span class="built_in">scanf</span>(<span class="string">"%lld%lld"</span>,&n,&mod)) {</span><br><span class="line"> <span class="function">Matrix <span class="title">a</span><span class="params">(<span class="number">3</span>)</span></span>;</span><br><span class="line"> a.A[<span class="number">1</span>][<span class="number">1</span>]=<span class="number">1</span>,a.A[<span class="number">1</span>][<span class="number">2</span>]=<span class="number">2</span>,a.A[<span class="number">1</span>][<span class="number">3</span>]=<span class="number">1</span>;</span><br><span class="line"> a.A[<span class="number">2</span>][<span class="number">1</span>]=<span class="number">1</span>,a.A[<span class="number">2</span>][<span class="number">2</span>]=<span class="number">1</span>,a.A[<span class="number">3</span>][<span class="number">1</span>]=<span class="number">1</span>;</span><br><span class="line"> Matrix ans=a^n;</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> A=ans.A[<span class="number">1</span>][<span class="number">1</span>],B=ans.A[<span class="number">2</span>][<span class="number">1</span>],C=ans.A[<span class="number">3</span>][<span class="number">1</span>];</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%lld\n"</span>,(A+mod-B+mod-B+C)%mod);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案-1"><a href="#答案-1" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">0</span></span><br><span class="line"><span class="number">1</span></span><br><span class="line"><span class="number">96</span></span><br><span class="line"><span class="number">64</span></span><br><span class="line"><span class="number">2503</span></span><br><span class="line"><span class="number">2523</span></span><br><span class="line"><span class="number">4452160</span></span><br><span class="line"><span class="number">557586868</span></span><br><span class="line"><span class="number">959316082</span></span><br><span class="line"><span class="number">1107500137</span></span><br></pre></td></tr></table></figure><h3 id="Program-3"><a href="#Program-3" class="headerlink" title="Program 3"></a>Program 3</h3><h4 id="分析-2"><a href="#分析-2" class="headerlink" title="分析"></a>分析</h4><p>容易发现：</p><script type="math/tex; mode=display">s_k=\sum_{i=0}^n i^k</script><p>其中任意数字的 $0$ 次方都定义为 $1$。</p><p>直接套用 $1,2,3,4$ 次方和公式即可。</p><script type="math/tex; mode=display">\begin{aligned}&s_0=n+1 \\&s_1=\frac{n(n+1)}{2} \\&s_2=\frac{n(n+1)(2n+1)}{6} \\&s_3=\frac{n^2(n+1)^2}{4} \\&s_4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\end{aligned}</script><h4 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> freopen(<span class="string">"program3.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program3.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> n;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%llu"</span>,&n);</span><br><span class="line"> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> s0=n+<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> s1=(n/<span class="number">2</span>)*(n+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> s2=(n/<span class="number">2</span>)*(n+<span class="number">1</span>)*((<span class="number">2</span>*n+<span class="number">1</span>)/<span class="number">3</span>);</span><br><span class="line"> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> s3=(n/<span class="number">2</span>)*(n/<span class="number">2</span>)*(n+<span class="number">1</span>)*(n+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> s4=(n/<span class="number">10</span>)*(n+<span class="number">1</span>)*((<span class="number">2</span>*n+<span class="number">1</span>)/<span class="number">3</span>)*(<span class="number">3</span>*n*n+<span class="number">3</span>*n<span class="number">-1</span>);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%llu\n%llu\n%llu\n%llu\n%llu\n%llu\n%llu\n%llu\n%llu\n%llu\n"</span>,s0,s0,s1,s1,s2,s2,s3,s3,s4,s4);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案-2"><a href="#答案-2" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">1000000000000001</span></span><br><span class="line"><span class="number">1000000000000001</span></span><br><span class="line"><span class="number">2538972135152631808</span></span><br><span class="line"><span class="number">2538972135152631808</span></span><br><span class="line"><span class="number">2806098670314569728</span></span><br><span class="line"><span class="number">2806098670314569728</span></span><br><span class="line"><span class="number">6570342264898322432</span></span><br><span class="line"><span class="number">6570342264898322432</span></span><br><span class="line"><span class="number">10067259324320137216</span></span><br><span class="line"><span class="number">10067259324320137216</span></span><br></pre></td></tr></table></figure><h3 id="Program-4"><a href="#Program-4" class="headerlink" title="Program 4"></a>Program 4</h3><h4 id="分析-3"><a href="#分析-3" class="headerlink" title="分析"></a>分析</h4><p>对于 $\text{count1}$ 函数，就是求有多少个无序点对满足两个点都是 $1$。直接统计 $1$ 点个数即可。</p><p>对于 $\text{count2}$ 函数，就是对于每个 $1$ 点，求出到它曼哈顿距离最近的白点的距离。可以直接从 $0$ 点 $\text{BFS}$。</p><h4 id="代码-3"><a href="#代码-3" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><queue></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">5e3</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> dx[]={<span class="number">1</span>,<span class="number">-1</span>,<span class="number">0</span>,<span class="number">0</span>},dy[]={<span class="number">0</span>,<span class="number">0</span>,<span class="number">1</span>,<span class="number">-1</span>};</span><br><span class="line"><span class="keyword">int</span> n,m,type,seed,dis[N][N];</span><br><span class="line"><span class="keyword">bool</span> a[N][N],vis[N][N];</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Node</span> {</span></span><br><span class="line"> <span class="keyword">int</span> x,y,w;</span><br><span class="line"> Node(<span class="keyword">int</span> _x=<span class="number">0</span>,<span class="keyword">int</span> _y=<span class="number">0</span>,<span class="keyword">int</span> _w=<span class="number">0</span>) {</span><br><span class="line"> x=_x,y=_y,w=_w;</span><br><span class="line"> }</span><br><span class="line">};</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">next_rand</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">static</span> <span class="keyword">const</span> <span class="keyword">int</span> P=<span class="number">1000000007</span>,Q=<span class="number">83978833</span>,R=<span class="number">8523467</span>;</span><br><span class="line"> <span class="keyword">return</span> seed=((<span class="keyword">long</span> <span class="keyword">long</span>)Q*seed%P*seed+R)%P;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">generate_input</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d"</span>,&n,&m,&type);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) a[i][j]=(next_rand()%<span class="number">8</span>><span class="number">0</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">count1</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) ans+=a[i][j];</span><br><span class="line"> <span class="keyword">return</span> ans*(ans<span class="number">-1</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">count2</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">queue</span><Node> q;</span><br><span class="line"> <span class="built_in">memset</span>(dis,<span class="number">0x3f</span>,<span class="keyword">sizeof</span>(dis));</span><br><span class="line"> <span class="built_in">memset</span>(vis,<span class="number">1</span>,<span class="keyword">sizeof</span>(vis));</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) {</span><br><span class="line"> <span class="keyword">if</span>(!a[i][j]) dis[i][j]=<span class="number">0</span>,vis[i][j]=<span class="number">1</span>,q.push(Node(i,j,<span class="number">0</span>));</span><br><span class="line"> <span class="keyword">else</span> vis[i][j]=<span class="number">0</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">while</span>(!q.empty()) {</span><br><span class="line"> Node u=q.front(); q.pop();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<<span class="number">4</span>;++i) {</span><br><span class="line"> <span class="keyword">int</span> x=u.x+dx[i],y=u.y+dy[i];</span><br><span class="line"> <span class="keyword">if</span>(vis[x][y]) <span class="keyword">continue</span>;</span><br><span class="line"> vis[x][y]=<span class="number">1</span>,dis[x][y]=u.w+<span class="number">1</span>,q.push(Node(x,y,u.w+<span class="number">1</span>));</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) ans+=dis[i][j];</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span>{</span><br><span class="line"> freopen(<span class="string">"program4.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program4.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&seed);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=<span class="number">10</span>;++i) {</span><br><span class="line"> generate_input();</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%lld\n"</span>,type==<span class="number">1</span>?count2():count1());</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案-3"><a href="#答案-3" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">65300</span></span><br><span class="line"><span class="number">768644095452</span></span><br><span class="line"><span class="number">1614752</span></span><br><span class="line"><span class="number">12299725860312</span></span><br><span class="line"><span class="number">6474661</span></span><br><span class="line"><span class="number">480452490358302</span></span><br><span class="line"><span class="number">40508992</span></span><br><span class="line"><span class="number">480453060258360</span></span><br><span class="line"><span class="number">40509116</span></span><br><span class="line"><span class="number">40508835</span></span><br></pre></td></tr></table></figure><h3 id="Program-5"><a href="#Program-5" class="headerlink" title="Program 5"></a>Program 5</h3><h4 id="分析-4"><a href="#分析-4" class="headerlink" title="分析"></a>分析</h4><p>源代码中的 $\text{count3}$ 函数就是在寻找有多少个全 $1$ 子矩阵。很显然是<strong>单调栈</strong>裸题。我们逐行分析，对于当前行维护每个点往上的最大全 $1$ 长度 $h_i$，求出 $h_i$ 左边第一个大于它的位置 $l_i$，右边第一个大于等于它的位置 $r_i$，那么对答案的贡献为 $h_i\times (i-l_i+1)(r_i-i+1)$。</p><h4 id="代码-4"><a href="#代码-4" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">5e3</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,seed,tp,h[N][N],l[N],r[N],st[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">next_rand</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">static</span> <span class="keyword">const</span> <span class="keyword">int</span> P=<span class="number">1000000007</span>,Q=<span class="number">83978833</span>,R=<span class="number">8523467</span>;</span><br><span class="line"> <span class="keyword">return</span> seed=((<span class="keyword">long</span> <span class="keyword">long</span>)Q*seed%P*seed+R)%P;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">calc</span><span class="params">(<span class="keyword">int</span> *a,<span class="keyword">int</span> n)</span> </span>{</span><br><span class="line"> st[tp=<span class="number">0</span>]=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="keyword">for</span>(;tp&&a[i]<=a[st[tp]];--tp);</span><br><span class="line"> l[i]=st[tp]+<span class="number">1</span>;</span><br><span class="line"> st[++tp]=i;</span><br><span class="line"> }</span><br><span class="line"> st[tp=<span class="number">0</span>]=n+<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=n;i>=<span class="number">1</span>;--i) {</span><br><span class="line"> <span class="keyword">for</span>(;tp&&a[i]<a[st[tp]];--tp);</span><br><span class="line"> r[i]=st[tp]<span class="number">-1</span>;</span><br><span class="line"> st[++tp]=i;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> ans+=<span class="number">1L</span>L*a[i]*(i-l[i]+<span class="number">1</span>)*(r[i]-i+<span class="number">1</span>);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">long</span> <span class="keyword">long</span> <span class="title">solve</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&m);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=m;++j) {</span><br><span class="line"> h[i][j]=(next_rand()%<span class="number">8</span>><span class="number">0</span>?h[i<span class="number">-1</span>][j]+<span class="number">1</span>:<span class="number">0</span>);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) ans+=calc(h[i],m);</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> freopen(<span class="string">"program5.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program5.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&seed);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<<span class="number">10</span>;++i) {</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%lld\n"</span>,solve());</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案-4"><a href="#答案-4" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">36798</span></span><br><span class="line"><span class="number">780109</span></span><br><span class="line"><span class="number">4970330</span></span><br><span class="line"><span class="number">19778353</span></span><br><span class="line"><span class="number">79444881</span></span><br><span class="line"><span class="number">183972917</span></span><br><span class="line"><span class="number">324090457</span></span><br><span class="line"><span class="number">401682783</span></span><br><span class="line"><span class="number">493647857</span></span><br><span class="line"><span class="number">493666110</span></span><br></pre></td></tr></table></figure><h3 id="Program-6"><a href="#Program-6" class="headerlink" title="Program 6"></a>Program 6</h3><h4 id="分析-5"><a href="#分析-5" class="headerlink" title="分析"></a>分析</h4><p>乍一看就是求 $f(f(f(\dots f(t))))$ 对 $c$ 取模，这个东西貌似是有环的？于是我们考虑找环，也就是找到这个<strong>环的起点</strong>。</p><p>我们可以使用 $\text{Floyd}$ <strong>判环算法</strong>，这个过程和 $\text{Pollard-Rho}$ 算法的实现很相似。因为这个环一定长成类似 $\rho$ 的形状。</p><p>我们设环长为 $n$，环之前的链长为 $m$，考虑使用<strong>两个指针</strong>，一个<strong>慢指针</strong>每次前进一步，一个<strong>快指针</strong>每次前进两步。两者都从起始点出发，当他们第一次相遇时一定有：</p><ul><li>慢指针移动了 $m+k_1 n+r$ 步。</li><li>快指针移动了 $m+k_2 n+r$ 步。</li></ul><p>由于快指针的速度是慢指针的两倍，那么设慢指针走了 $S$ 步，快指针走了 $2S$ 步，两式相减得到：</p><script type="math/tex; mode=display">S=(k_2-k_1)n</script><p>这意味着他们走过的路程都是<strong>环长的若干倍</strong>。</p><p>接下来我们把慢指针移动到起始位置，让两个指针都只移动一步，当慢指针移动了 $m$ 步，快指针总共移动了 $2S+m$ 步。对这个 $2S+m$ 有如下分析：</p><ol><li>计算出这个值：快指针和慢指针速度相同，都移动了 $m$ 的距离，加上之前的 $2S$ 就是 $2S+m$ 步。</li><li>换个角度思考：从起始位置走 $m$ 步到达环的起点，由于 $2S$ 是环长的整数倍，那么一定又走到了环的起点！</li></ol><p>这样一来我们就得到了环的起点！</p><p>当然本题可以用更加优秀的 $\text{Brent}$ 算法。</p><p><del>这个环长貌似是期望 $\sqrt c$ 的代？码只要十分钟就可以跑出来了！</del></p><h4 id="代码-5"><a href="#代码-5" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><iostream></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="keyword">typedef</span> <span class="keyword">unsigned</span> <span class="keyword">long</span> <span class="keyword">long</span> ull;</span><br><span class="line">ull n,a,b,c;</span><br><span class="line"></span><br><span class="line"><span class="function">ull <span class="title">F</span><span class="params">(ull x)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> (x*x*a+b)%c;</span><br><span class="line">}</span><br><span class="line"><span class="function">ull <span class="title">calcChain</span><span class="params">(ull x,ull &y)</span> </span>{</span><br><span class="line"> ull len=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(x!=y) x=F(x),y=F(F(y)),++len;</span><br><span class="line"> <span class="keyword">return</span> len;</span><br><span class="line">}</span><br><span class="line"><span class="function">ull <span class="title">calcLoop</span><span class="params">(ull x)</span> </span>{</span><br><span class="line"> ull st=x,len=<span class="number">1</span>;</span><br><span class="line"> x=F(x);</span><br><span class="line"> <span class="keyword">while</span>(x!=st) x=F(x),++len;</span><br><span class="line"> <span class="keyword">return</span> len;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">()</span> </span>{</span><br><span class="line"> ull x=F(<span class="number">0</span>),y=F(F(<span class="number">0</span>));</span><br><span class="line"> <span class="keyword">while</span>(x!=y) x=F(x),y=F(F(y));</span><br><span class="line"> ull m=calcChain(<span class="number">0</span>,y);</span><br><span class="line"> ull len=calcLoop(y);</span><br><span class="line"> <span class="keyword">if</span>(n<m) {</span><br><span class="line"> ull ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(ull i=<span class="number">1</span>;i<=n;++i) ans=F(ans);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%llu\n"</span>,ans);</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> n-=m;</span><br><span class="line"> ull ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(ull i=<span class="number">1</span>;i<=m;++i) ans=F(ans);</span><br><span class="line"> <span class="keyword">for</span>(ull i=<span class="number">1</span>;i<=n%len;++i) ans=F(ans);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%llu\n"</span>,ans);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span>;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> freopen(<span class="string">"program6.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program6.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<<span class="number">10</span>;++i) {</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%llu%llu%llu%llu"</span>,&n,&a,&b,&c);</span><br><span class="line"> solve();</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案-5"><a href="#答案-5" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">518048760868869048</span></span><br><span class="line"><span class="number">1792066212437514363</span></span><br><span class="line"><span class="number">31017889126204134</span></span><br><span class="line"><span class="number">2107151525961152753</span></span><br><span class="line"><span class="number">402987993063476955</span></span><br><span class="line"><span class="number">1026935915030784632</span></span><br><span class="line"><span class="number">2533709394548916391</span></span><br><span class="line"><span class="number">1357484894607415330</span></span><br><span class="line"><span class="number">1099871450072879095</span></span><br><span class="line"><span class="number">235900336338336004</span></span><br></pre></td></tr></table></figure><h3 id="Program-7"><a href="#Program-7" class="headerlink" title="Program 7"></a>Program 7</h3><h4 id="分析-6"><a href="#分析-6" class="headerlink" title="分析"></a>分析</h4><p>完成一个 $16\times 16$ 的数独<del>（字母独）</del>。</p><p>加上一些剪枝可以轻松跑出来前几个数独，最后几个数独需要采用<strong>倒搜</strong>。</p><h4 id="代码-6"><a href="#代码-6" class="headerlink" title="代码"></a>代码</h4><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">20</span>;</span><br><span class="line"><span class="keyword">char</span> s[N][N];</span><br><span class="line"><span class="keyword">bool</span> row[N][N],col[N][N],seq[N][N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">getSeq</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> r=x/<span class="number">4</span>+<span class="number">1</span>,c=y/<span class="number">4</span>+<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">return</span> (r<span class="number">-1</span>)*<span class="number">4</span>+c;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">bool</span> <span class="title">dfs</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> <span class="keyword">if</span>(x<<span class="number">0</span>) <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line"> <span class="keyword">if</span>(s[x][y]!=<span class="string">'?'</span>) <span class="keyword">return</span> y==<span class="number">0</span>?dfs(x<span class="number">-1</span>,<span class="number">15</span>):dfs(x,y<span class="number">-1</span>);</span><br><span class="line"> <span class="keyword">int</span> z=getSeq(x,y);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">char</span> i=<span class="string">'A'</span>;i<=<span class="string">'P'</span>;++i) {</span><br><span class="line"> <span class="keyword">if</span>(row[x][i-<span class="string">'A'</span>]||col[y][i-<span class="string">'A'</span>]||seq[z][i-<span class="string">'A'</span>]) <span class="keyword">continue</span>;</span><br><span class="line"> row[x][i-<span class="string">'A'</span>]=col[y][i-<span class="string">'A'</span>]=seq[z][i-<span class="string">'A'</span>]=<span class="number">1</span>;</span><br><span class="line"> s[x][y]=i;</span><br><span class="line"> <span class="keyword">if</span>(y==<span class="number">0</span>?dfs(x<span class="number">-1</span>,<span class="number">15</span>):dfs(x,y<span class="number">-1</span>)) <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line"> row[x][i-<span class="string">'A'</span>]=col[y][i-<span class="string">'A'</span>]=seq[z][i-<span class="string">'A'</span>]=<span class="number">0</span>;</span><br><span class="line"> s[x][y]=<span class="string">'?'</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">(<span class="keyword">int</span> points)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<=<span class="number">15</span>;++i) <span class="built_in">scanf</span>(<span class="string">"%s"</span>,s[i]);</span><br><span class="line"> <span class="built_in">memset</span>(row,<span class="number">0</span>,<span class="keyword">sizeof</span>(row));</span><br><span class="line"> <span class="built_in">memset</span>(col,<span class="number">0</span>,<span class="keyword">sizeof</span>(col));</span><br><span class="line"> <span class="built_in">memset</span>(seq,<span class="number">0</span>,<span class="keyword">sizeof</span>(seq));</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<=<span class="number">15</span>;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>;j<=<span class="number">15</span>;++j) <span class="keyword">if</span>(s[i][j]!=<span class="string">'?'</span>) {</span><br><span class="line"> row[i][s[i][j]-<span class="string">'A'</span>]=<span class="number">1</span>,col[j][s[i][j]-<span class="string">'A'</span>]=<span class="number">1</span>,seq[getSeq(i,j)][s[i][j]-<span class="string">'A'</span>]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(dfs(<span class="number">15</span>,<span class="number">15</span>)) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">1</span>;k<=points;++k,<span class="built_in">puts</span>(<span class="string">""</span>)) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<=<span class="number">15</span>;++i) <span class="built_in">printf</span>(<span class="string">"%s"</span>,s[i]);</span><br><span class="line"> }</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">1</span>;k<=points;++k) <span class="built_in">puts</span>(<span class="string">"NO SOLUTION"</span>);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> freopen(<span class="string">"program7.in"</span>,<span class="string">"r"</span>,<span class="built_in">stdin</span>);</span><br><span class="line"> freopen(<span class="string">"program7.out"</span>,<span class="string">"w"</span>,<span class="built_in">stdout</span>);</span><br><span class="line"> solve(<span class="number">1</span>);</span><br><span class="line"> solve(<span class="number">2</span>);</span><br><span class="line"> solve(<span class="number">3</span>);</span><br><span class="line"> solve(<span class="number">4</span>);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h4 id="答案-6"><a href="#答案-6" class="headerlink" title="答案"></a>答案</h4><figure class="highlight armasm"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="symbol">NAPILFJBMHEOGDCKJDCHMNIPAFGKOBLEKLMGAEHONCDBPJFIFEOBCDGKIJLPAMHNOFHLJMBGEKIDCNAPPCEAHLDFBMJNIKGOBNIKOAPEFGCHJLMDMGDJNIKCLOPAFHEBIOBPFKLJHEMGDANCDKJEIGMAOBNCLFPHGMNCEPOHDLAFKIBJAHLFDBCNPIKJEGOMEIFMBHADKPOLNCJGHPADGJFICNBEMOKLLJKOPCNMGAHIBEDFCBGNKOELJDFMHPIA</span></span><br><span class="line"><span class="keyword">BLNOMGAPIHJKDCEFHGIMCFBOAPEDKNLJFCAEDKLJBGNOPHMIDKJPHIENFMLCGBAOIBEAKCFLHNOGJMDPLHFNJAPDCIKMEGOBPDOJNMHGELBFAKICCMGKOBIEJADPFLNHOFCHPNMBKJAELIGDJNLDGECHOBFIMPKAGAPIFDOKLCMNHJBEMEKBALJIGDPHOFCNEIBGLPNMDOHJCAFKKPDCEHGANFILBOJMNJHLBOKFMECAIDPGAOMFIJDCPKGBNEHL</span></span><br><span class="line"><span class="keyword">BLNOMGAPIHJKDCEFHGIMCFBOAPEDKNLJFCAEDKLJBGNOPHMIDKJPHIENFMLCGBAOIBEAKCFLHNOGJMDPLHFNJAPDCIKMEGOBPDOJNMHGELBFAKICCMGKOBIEJADPFLNHOFCHPNMBKJAELIGDJNLDGECHOBFIMPKAGAPIFDOKLCMNHJBEMEKBALJIGDPHOFCNEIBGLPNMDOHJCAFKKPDCEHGANFILBOJMNJHLBOKFMECAIDPGAOMFIJDCPKGBNEHL</span></span><br><span class="line"><span class="keyword">FMPCJGLEANDBOHKIOEHGNCAMFKILDJPBDBNIFHKPJOGEMACLKALJOIDBCHMPGNEFEGOHPJBKIFNMADLCALFDIENOHJKCBMGPNKCPMFGDBLEAHIJOJIBMHACLODPGEFNKLDIEAPHGNBCOJKFMMHAODLJCPGFKNBIEGPKFEBMNDIAJLCOHBCJNKOFIMELHPGADIOGKCDEHLABNFPMJPFEBLMIAGCJDKOHNHJDLGNPFKMOICEBACNMABKOJEPHFILDG</span></span><br><span class="line"><span class="keyword">FMPCJGLEANDBOHKIOEHGNCAMFKILDJPBDBNIFHKPJOGEMACLKALJOIDBCHMPGNEFEGOHPJBKIFNMADLCALFDIENOHJKCBMGPNKCPMFGDBLEAHIJOJIBMHACLODPGEFNKLDIEAPHGNBCOJKFMMHAODLJCPGFKNBIEGPKFEBMNDIAJLCOHBCJNKOFIMELHPGADIOGKCDEHLABNFPMJPFEBLMIAGCJDKOHNHJDLGNPFKMOICEBACNMABKOJEPHFILDG</span></span><br><span class="line"><span class="keyword">FMPCJGLEANDBOHKIOEHGNCAMFKILDJPBDBNIFHKPJOGEMACLKALJOIDBCHMPGNEFEGOHPJBKIFNMADLCALFDIENOHJKCBMGPNKCPMFGDBLEAHIJOJIBMHACLODPGEFNKLDIEAPHGNBCOJKFMMHAODLJCPGFKNBIEGPKFEBMNDIAJLCOHBCJNKOFIMELHPGADIOGKCDEHLABNFPMJPFEBLMIAGCJDKOHNHJDLGNPFKMOICEBACNMABKOJEPHFILDG</span></span><br><span class="line"><span class="keyword">LDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPN</span></span><br><span class="line"><span class="keyword">LDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPN</span></span><br><span class="line"><span class="keyword">LDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPN</span></span><br><span class="line"><span class="keyword">LDFABIOEGCKJNMHPJPONCFDMLHABIGKEHMEGLJPKFNOIACDBCIKBANHGPMEDFOJLFBMIJPLHAKGEDNCOACJKODIFNLPHBEMGPGHEMBKNDFCOJILAONLDEAGCIJBMPHFKDJIPHOELCAFNGKBMEOGHPMCABIJKLDNFMFACNKJBOGDLEPIHBKNLIGFDMEHPCAOJILCJGHMPKBNAOFEDGHBFDLNOEPMCKJAINAPOKEBJHDIFMLGCKEDMFCAIJOLGHBPN</span></span><br></pre></td></tr></table></figure><h3 id="Program-8"><a href="#Program-8" class="headerlink" title="Program 8"></a>Program 8</h3><h4 id="分析-7"><a href="#分析-7" class="headerlink" title="分析"></a>分析</h4><p>我们发现这个代码本质就是求，对于所有的整数 $a,b,c,d,e,f,g\in [1,n]$，满足各种奇淫不等式的七元组有几个？</p><p>由于有 $7$ 个未知数，那么答案一定是关于 $n$ 的一个 $7$ 次多项式。可以暴力跑出来 $n\in [1,8]$ 的答案，然后<strong>拉格朗日插值</strong>即可。</p><h4 id="代码-7"><a href="#代码-7" class="headerlink" title="代码"></a>代码</h4><p>此处不放出具体代码了，拉格朗日插值的写法详见<a href="https://orzsiyuan.com/articles/algorithm-Lagrange-Interpolation" target="_blank" rel="noopener">「算法笔记」拉格朗日插值</a>。</p><h4 id="答案-7"><a href="#答案-7" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">1018333390</span></span><br><span class="line"><span class="number">993704934</span></span><br><span class="line"><span class="number">1053807588</span></span><br><span class="line"><span class="number">1144151985</span></span><br><span class="line"><span class="number">712062141</span></span><br><span class="line"><span class="number">530076748</span></span><br><span class="line"><span class="number">520686243</span></span><br><span class="line"><span class="number">337499021</span></span><br><span class="line"><span class="number">820275783</span>同询问 $<span class="number">5</span>$，`hold`。</span><br><span class="line"><span class="number">80253986</span></span><br></pre></td></tr></table></figure><h3 id="Program-9"><a href="#Program-9" class="headerlink" title="Program 9"></a>Program 9</h3><h3 id="分析-8"><a href="#分析-8" class="headerlink" title="分析"></a>分析</h3><ol><li>答案为 $1984$。</li><li>常用密码 $123456$。</li><li><del>我怎么知道这个人是</del> <code>chenlijie</code>。</li><li>爆搜可以得到 <script type="math/tex">_</script>。</li><li>发现在 Program $2$ 中有一部伪英语字典，将其中每个单词的 $\text{MD5}$ 跑出来后和输入文件中的对比可以得到 <code>we</code>。</li><li>同询问 $5$，<code>hold</code>。</li><li>同询问 $5$，<code>these</code>。</li><li>同询问 $5$，<code>truths</code>。</li><li>由于最后一行的提示 <code>最后 6 行组成了一句名言</code>，可以在 Program $10$ 中找到这句话 <code>We hold these truths to be selfevident</code>。因此为 <code>to be</code>。</li><li>同询问 $9$，<code>selfevident</code>。</li></ol><h4 id="代码-8"><a href="#代码-8" class="headerlink" title="代码"></a>代码</h4><p><del>要什么代码？QAQ</del></p><h4 id="答案-8"><a href="#答案-8" class="headerlink" title="答案"></a>答案</h4><figure class="highlight autoit"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">1984</span></span><br><span class="line"><span class="number">123456</span></span><br><span class="line">chenlijie</span><br><span class="line">$_$</span><br><span class="line">we</span><br><span class="line">hold</span><br><span class="line">these</span><br><span class="line">truths</span><br><span class="line"><span class="keyword">to</span> be</span><br><span class="line">selfevident</span><br></pre></td></tr></table></figure><h3 id="Program-10"><a href="#Program-10" class="headerlink" title="Program 10"></a>Program 10</h3><h4 id="分析-9"><a href="#分析-9" class="headerlink" title="分析"></a>分析</h4><p>发现每个单词的函数运行时间不长，而瓶颈主要在 <code>A</code> 到 <code>Z</code> 函数。而他们的本质就是执行若干次 <code>_</code> 函数，直接<strong>递推</strong>出这些函数分别执行了多少次 <code>_</code> 即可。</p><h4 id="代码-9"><a href="#代码-9" class="headerlink" title="代码"></a>代码</h4><p><del>这么长代码放上来干嘛？QAQ</del></p><h4 id="答案-9"><a href="#答案-9" class="headerlink" title="答案"></a>答案</h4><figure class="highlight lsl"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="number">6754098618987872142</span></span><br><span class="line"><span class="number">3227259651709640332</span></span><br><span class="line"><span class="number">3227259651709640332</span></span><br><span class="line"><span class="number">1514994897561930504</span></span><br><span class="line"><span class="number">1514994897561930504</span></span><br><span class="line"><span class="number">9307175031877652086</span></span><br><span class="line"><span class="number">9307175031877652086</span></span><br><span class="line"><span class="number">9307175031877652086</span></span><br><span class="line"><span class="number">9307175031877652086</span></span><br><span class="line"><span class="number">9307175031877652086</span></span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="http://uoj.ac/problem/73" target="_blank" rel="noopener">UOJ 73</a></p>
</blockquote>
<p>本题是一道提交答案题，一共有 $10$ 个测试点。</p>
<p>对于每个测试点，你会得到一段程序的源代码和这段程序的输入。你要运行这个程序，并保存这个程序的输出。</p>
<p>遗憾的是这些程序效率都极其低下，无法在比赛的 $5$ 小时内得到输出。</p>
<p>你需要帮助 B 君得到这些程序的输出。</p>
<p><a href="http://uoj.ac/download.php?type=problem&amp;id=73" target="_blank" rel="noopener">输入数据下载</a></p>
</summary>
<category term="拉格朗日插值" scheme="http://hydingsy.github.io/tags/%E6%8B%89%E6%A0%BC%E6%9C%97%E6%97%A5%E6%8F%92%E5%80%BC/"/>
<category term="UOJ" scheme="http://hydingsy.github.io/tags/UOJ/"/>
<category term="单调栈" scheme="http://hydingsy.github.io/tags/%E5%8D%95%E8%B0%83%E6%A0%88/"/>
<category term="BFS" scheme="http://hydingsy.github.io/tags/BFS/"/>
<category term="矩阵快速幂" scheme="http://hydingsy.github.io/tags/%E7%9F%A9%E9%98%B5%E5%BF%AB%E9%80%9F%E5%B9%82/"/>
<category term="WC" scheme="http://hydingsy.github.io/tags/WC/"/>
<category term="提交答案题" scheme="http://hydingsy.github.io/tags/%E6%8F%90%E4%BA%A4%E7%AD%94%E6%A1%88%E9%A2%98/"/>
<category term="Floyd 判环算法" scheme="http://hydingsy.github.io/tags/Floyd-%E5%88%A4%E7%8E%AF%E7%AE%97%E6%B3%95/"/>
</entry>
<entry>
<title>「算法笔记」拉格朗日插值</title>
<link href="http://hydingsy.github.io/articles/algorithm-Lagrange-Interpolation/"/>
<id>http://hydingsy.github.io/articles/algorithm-Lagrange-Interpolation/</id>
<published>2019-02-25T16:00:00.000Z</published>
<updated>2019-03-03T02:16:45.548Z</updated>
<content type="html"><![CDATA[<blockquote class="blockquote-center">如果对实践中的某个物理量进行观测，在若干个不同的地方得到相应的观测值，拉格朗日插值法可以找到一个多项式，其恰好在各个观测的点取到观测到的值。</blockquote><a id="more"></a><h2 id="概述"><a href="#概述" class="headerlink" title="概述"></a>概述</h2><p>众所周知，平面上 $n+1$ 个点可以确定一个 $n$ 次多项式 $P(x)$。现在已知 $n+1$ 个点 $(x_i,y_i)$，请确定经过这 $n+1$ 个点的多项式 $P(x)$ 在 $x=k$ 时的值。</p><p>拉格朗日插值可以解决这类问题！</p><hr><h2 id="实现"><a href="#实现" class="headerlink" title="实现"></a>实现</h2><p>拉格朗日插值的经精妙之处就在于<strong>拉格朗日基本多项式</strong>。</p><p>我们设这 $n$ 个点为 $(x_0,y_0),(x_1,y_1),\dots,(x_n,y_n)$。那么我们构造多项式：</p><script type="math/tex; mode=display">\ell_i(x)=\prod_{j=0,i\neq j}^n \frac{k-x_j}{x_i-x_j}</script><p>这个多项式的构造十分巧妙，我们可以注意到如下规律：</p><script type="math/tex; mode=display">\ell_i(x_j)=\begin{cases}1 & i=j \\0 & i\neq j\end{cases}</script><p>接着我们构造这个 $n$ 次多项式：</p><script type="math/tex; mode=display">P(x)=\sum_{i=0}^n y_i\ell_i(x)</script><p>根据基本多项式的性质，我们可以得到 $P(x_i)=y_i$，也就是说保证经过这 $n+1$ 个点，经过简单的计算就可以得到<strong>系数表达式</strong>，当然也很容易得到 $P(k)$ 的值了。</p><p><strong>时间复杂度</strong>：$O(n^2)$</p><hr><h2 id="拓展"><a href="#拓展" class="headerlink" title="拓展"></a>拓展</h2><h3 id="取值连续"><a href="#取值连续" class="headerlink" title="取值连续"></a>取值连续</h3><p>考虑在 $x$ 的值连续的情况下，如何快速求 $P(k)$ 的值？（这里的 $x$ 的值连续的意思是：对于 $0\le i\le n$，$x_i=i$）</p><p>那么式子化为：</p><script type="math/tex; mode=display">P(x)=\sum_{i=0}^n y_i\prod_{j=0,i\neq j}^n \frac{x-j}{i-j}</script><p>对于分子，我们维护 $pre_i=\prod_{j=0}^i k-j$，$suf_i=\prod_{j=i}^n k-j$；而对于分母，我们发现就是阶乘的形式！于是这个式子变成了：</p><script type="math/tex; mode=display">P(k)=\sum_{i=0}^n y_i\frac{pre_{i-1}\cdot suf_{i+1}}{i!\cdot (n-i)!}</script><p>但是注意：分母可能出现符号问题，也就是说说当 $n-i$ 为负数时，分母应该取<strong>负号</strong>！</p><p><strong>时间复杂度</strong>：$O(n)$</p><h3 id="经典应用"><a href="#经典应用" class="headerlink" title="经典应用"></a>经典应用</h3><p>求解 $1$ 到 $n$ 的 $k$ 次方和。即如下式子：</p><script type="math/tex; mode=display">\sum_{i=1}^n i^k</script><p>数据范围：$1\le n\le 10^{18}$，$1\le k\le 10^6$</p><p>这个东西显然是一个以 $n$ 为自变量的 $k+1$ 次多项式。那么我们代入 $1\sim k+2$ 这一共 $k+2$ 个点后直接拉格朗日插值即可！</p><blockquote><p><strong>定理</strong>：若多项式 $P(x)$ 的次数为 $n$，那么对于任何 $k\ge n+1$，多项式 $P(x)$ 的 $k$ 阶差分恒等于 $0$。</p></blockquote><p>我们预处理出 $pre_i$ 和 $suf_i$，在处理 $k+2$ 个点时可以采用<strong>线性筛</strong>求出每个数的 $k$ 次方。</p><p><strong>时间复杂度</strong>：$O(k)$</p><hr><h2 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h2><p>我们用一个函数 $\text{lagrange}$ 来实现。其中 <code>*x,*y</code> 分别存点的坐标，函数的返回值为 $P(k)$ 的值（其中 $\text{mod}$ 为模数，$\text{inv}(x)$ 表示 $x$ 的逆元，$\text{upd}(x,y)$ 表示将 $x$ 加上 $y$）。</p><h3 id="常用写法"><a href="#常用写法" class="headerlink" title="常用写法"></a>常用写法</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">lagrange</span><span class="params">(<span class="keyword">int</span> n,<span class="keyword">int</span> *x,<span class="keyword">int</span> *y,<span class="keyword">int</span> k)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="keyword">int</span> f=<span class="number">1</span>,g=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) <span class="keyword">if</span>(i!=j) {</span><br><span class="line"> f=<span class="number">1L</span>L*f*(k-x[j]+mod)%mod;</span><br><span class="line"> g=<span class="number">1L</span>L*g*(x[i]-x[j]+mod)%mod;</span><br><span class="line"> }</span><br><span class="line"> upd(ans,<span class="number">1L</span>L*y[i]*f%mod*inv(g)%mod);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h3 id="取值连续-1"><a href="#取值连续-1" class="headerlink" title="取值连续"></a>取值连续</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">lagrange</span><span class="params">(<span class="keyword">int</span> n,<span class="keyword">int</span> *y,<span class="keyword">int</span> k)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> pre[<span class="number">0</span>]=suf[n+<span class="number">1</span>]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) pre[i]=<span class="number">1L</span>L*pre[i<span class="number">-1</span>]*(k-i)%mod;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=n;i>=<span class="number">1</span>;--i) suf[i]=<span class="number">1L</span>L*suf[i+<span class="number">1</span>]*(k-i)%mod;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="keyword">int</span> a=<span class="number">1L</span>L*pre[i<span class="number">-1</span>]*suf[i+<span class="number">1</span>]%mod*inv[i<span class="number">-1</span>]%mod*inv[n-i]%mod;</span><br><span class="line"> <span class="keyword">if</span>((n-i)&<span class="number">1</span>) a=mod-a;</span><br><span class="line"> upd(ans,<span class="number">1L</span>L*a*y[i]%mod);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><hr><h2 id="习题"><a href="#习题" class="headerlink" title="习题"></a>习题</h2><ul><li><a href="https://www.luogu.org/problemnew/show/P4781" target="_blank" rel="noopener">「Luogu 4781」【模板】拉格朗日插值</a></li><li><a href="http://www.51nod.com/Challenge/Problem.html#!#problemId=1228" target="_blank" rel="noopener">「51nod 1228」序列求和 V1</a></li><li><a href="http://www.51nod.com/Challenge/Problem.html#!#problemId=1258" target="_blank" rel="noopener">「51nod 1258」序列求和 V4</a></li><li><a href="https://www.lydsy.com/JudgeOnline/problem.php?id=5339" target="_blank" rel="noopener">「TJOI 2018」教科书般的亵渎</a></li><li><a href="https://www.lydsy.com/JudgeOnline/problem.php?id=3453" target="_blank" rel="noopener">「BZOJ 3453」XLKxc</a></li><li><a href="https://www.lydsy.com/JudgeOnline/problem.php?id=4559" target="_blank" rel="noopener">「JLOI 2016」成绩比较</a></li><li><a href="https://www.lydsy.com/JudgeOnline/problem.php?id=2655" target="_blank" rel="noopener">「BZOJ 2655」Calc</a></li></ul>]]></content>
<summary type="html">
<blockquote class="blockquote-center">如果对实践中的某个物理量进行观测，在若干个不同的地方得到相应的观测值，拉格朗日插值法可以找到一个多项式，其恰好在各个观测的点取到观测到的值。</blockquote>
</summary>
<category term="算法笔记" scheme="http://hydingsy.github.io/tags/%E7%AE%97%E6%B3%95%E7%AC%94%E8%AE%B0/"/>
<category term="数论" scheme="http://hydingsy.github.io/tags/%E6%95%B0%E8%AE%BA/"/>
<category term="拉格朗日插值" scheme="http://hydingsy.github.io/tags/%E6%8B%89%E6%A0%BC%E6%9C%97%E6%97%A5%E6%8F%92%E5%80%BC/"/>
</entry>
<entry>
<title>「Codeforces 1097D」Makoto and a Blackboard</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-1097D-Makoto-and-a-Blackboard/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-1097D-Makoto-and-a-Blackboard/</id>
<published>2019-02-24T16:00:00.000Z</published>
<updated>2019-02-28T04:04:06.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/1097/problem/D" target="_blank" rel="noopener">Codeforces 1097D</a></p></blockquote><p>Makoto 有一个大黑板，上面写着一个正整数 $n$，他将会进行恰好 $k$ 次操作：</p><p>假设黑板上的数字是 $v$，他会随机选择一个 $v$ 的约数（可能是 $1$ 或 $v$）然后用这个约数替换 $v$。保证每个约数被选择的概率相同。</p><p>他想知道 $k$ 次操作后黑板上的数字的期望值是多少。答案对 $10^9+7$ 取模。</p><p>数据范围：$1\le n\le 10^{15}$，$1\le k\le 10^4$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>我们定义 $f(i,j)$ 表示操作 $i$ 次后得到数字 $j$ 的概率。发现每个质因子都是<strong>相互独立</strong>的，每次改变的只不过是若干个<strong>质因子的指数</strong>。那么这意味着 $f$ 是一个<strong>积性函数</strong>！</p><p>考虑对于 $n$ 的一个质因子 $p$，其指数为 $c$，那么设 $g(i,j)$ 表示数字 $p^c$ 操作 $i$ 次后指数为 $j$ 的概率。转移方程为：</p><script type="math/tex; mode=display">g(i,j)=\sum_{k=j}^c \frac{g(i-1,k)}{k+1}</script><p>这个式子的含义是：指数 $j$ 可以由指数 $j\sim c$ 转移过来，对于原来的指数 $k$，有 $\frac{1}{k+1}$ 的概率转移到现在的指数 $j$（其实这个式子可以<strong>前缀和优化</strong>，复杂度由 $O(kc^2)$ 优化为 $O(kc)$，但是由于 $c\le \log n$，因此无关紧要）。</p><p>那么数字 $p^c$ 操作后的期望为：</p><script type="math/tex; mode=display">\sum_{i=0}^c g(k,i)\times p^i</script><p>最终的答案为所有质因子答案的<strong>乘积</strong>（积性函数的定义）。</p><p><strong>时间复杂度</strong>：$O(k\log^3 n)$（非常不满），优化后为 $O(k\log^2 n)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><h3 id="朴素写法"><a href="#朴素写法" class="headerlink" title="朴素写法"></a>朴素写法</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><vector></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> K=<span class="number">1e4</span>+<span class="number">5</span>,logN=<span class="number">51</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> mod=<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> n;</span><br><span class="line"><span class="keyword">int</span> k,f[K][logN],inv[logN];</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">vector</span><<span class="built_in">std</span>::pair<<span class="keyword">int</span>,<span class="keyword">int</span>> > v;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">getFactor</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> n)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;<span class="number">1L</span>L*i*i<=n;++i) {</span><br><span class="line"> <span class="keyword">if</span>(n%i) <span class="keyword">continue</span>;</span><br><span class="line"> <span class="keyword">int</span> cnt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(n%i==<span class="number">0</span>) n/=i,++cnt;</span><br><span class="line"> v.push_back(<span class="built_in">std</span>::make_pair(i,cnt));</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(n><span class="number">1</span>) v.push_back(<span class="built_in">std</span>::make_pair(n%mod,<span class="number">1</span>));</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">initInv</span><span class="params">()</span> </span>{</span><br><span class="line"> inv[<span class="number">0</span>]=inv[<span class="number">1</span>]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i<logN;++i) inv[i]=<span class="number">1L</span>L*(mod-mod/i)*inv[mod%i]%mod;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">pow</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> p)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(;p;p>>=<span class="number">1</span>,x=<span class="number">1L</span>L*x*x%mod) <span class="keyword">if</span>(p&<span class="number">1</span>) ans=<span class="number">1L</span>L*ans*x%mod;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">upd</span><span class="params">(<span class="keyword">int</span> &x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> (x+=y)>=mod&&(x-=mod);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">solve</span><span class="params">(<span class="keyword">int</span> p,<span class="keyword">int</span> c)</span> </span>{</span><br><span class="line"> <span class="built_in">memset</span>(f,<span class="number">0</span>,<span class="keyword">sizeof</span>(f));</span><br><span class="line"> f[<span class="number">0</span>][c]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=k;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>;j<=c;++j) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> l=j;l<=c;++l) upd(f[i][j],<span class="number">1L</span>L*f[i<span class="number">-1</span>][l]*inv[l+<span class="number">1</span>]%mod);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<=c;++i) upd(ans,<span class="number">1L</span>L*f[k][i]*<span class="built_in">pow</span>(p,i)%mod);</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld%d"</span>,&n,&k);</span><br><span class="line"> getFactor(n);</span><br><span class="line"> initInv();</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="built_in">std</span>::pair<<span class="keyword">int</span>,<span class="keyword">int</span>> x:v) ans=<span class="number">1L</span>L*ans*solve(x.first,x.second)%mod;</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure><h3 id="前缀和优化"><a href="#前缀和优化" class="headerlink" title="前缀和优化"></a>前缀和优化</h3><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><vector></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> logN=<span class="number">51</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> mod=<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"><span class="keyword">long</span> <span class="keyword">long</span> n;</span><br><span class="line"><span class="keyword">int</span> k,f[logN],inv[logN],s[logN];</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">vector</span><<span class="built_in">std</span>::pair<<span class="keyword">int</span>,<span class="keyword">int</span>> > v;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">getFactor</span><span class="params">(<span class="keyword">long</span> <span class="keyword">long</span> n)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;<span class="number">1L</span>L*i*i<=n;++i) {</span><br><span class="line"> <span class="keyword">if</span>(n%i) <span class="keyword">continue</span>;</span><br><span class="line"> <span class="keyword">int</span> cnt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(n%i==<span class="number">0</span>) n/=i,++cnt;</span><br><span class="line"> v.push_back(<span class="built_in">std</span>::make_pair(i,cnt));</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(n><span class="number">1</span>) v.push_back(<span class="built_in">std</span>::make_pair(n%mod,<span class="number">1</span>));</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">initInv</span><span class="params">()</span> </span>{</span><br><span class="line"> inv[<span class="number">0</span>]=inv[<span class="number">1</span>]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i<logN;++i) inv[i]=<span class="number">1L</span>L*(mod-mod/i)*inv[mod%i]%mod;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">pow</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> p)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(;p;p>>=<span class="number">1</span>,x=<span class="number">1L</span>L*x*x%mod) <span class="keyword">if</span>(p&<span class="number">1</span>) ans=<span class="number">1L</span>L*ans*x%mod;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">upd</span><span class="params">(<span class="keyword">int</span> &x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> (x+=y)>=mod&&(x-=mod);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">solve</span><span class="params">(<span class="keyword">int</span> p,<span class="keyword">int</span> c)</span> </span>{</span><br><span class="line"> <span class="built_in">memset</span>(f,<span class="number">0</span>,<span class="keyword">sizeof</span>(f)),<span class="built_in">memset</span>(s,<span class="number">0</span>,<span class="keyword">sizeof</span>(s));</span><br><span class="line"> f[c]=<span class="number">1</span>,s[c]=f[c]*inv[c+<span class="number">1</span>];</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=k;++i) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>;j<=c;++j) f[j]=(s[c]-(j?s[j<span class="number">-1</span>]:<span class="number">0</span>)+mod)%mod;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>;j<=c;++j) s[j]=<span class="number">1L</span>L*f[j]*inv[j+<span class="number">1</span>]%mod;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=c;++j) upd(s[j],s[j<span class="number">-1</span>]);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<=c;++i) upd(ans,<span class="number">1L</span>L*f[i]*<span class="built_in">pow</span>(p,i)%mod);</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld%d"</span>,&n,&k);</span><br><span class="line"> getFactor(n);</span><br><span class="line"> initInv();</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="built_in">std</span>::pair<<span class="keyword">int</span>,<span class="keyword">int</span>> x:v) ans=<span class="number">1L</span>L*ans*solve(x.first,x.second)%mod;</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/1097/problem/D" target="_blank" rel="noopener">Codeforces 1097D</a></p>
</blockquote>
<p>Makoto 有一个大黑板，上面写着一个正整数 $n$，他将会进行恰好 $k$ 次操作：</p>
<p>假设黑板上的数字是 $v$，他会随机选择一个 $v$ 的约数（可能是 $1$ 或 $v$）然后用这个约数替换 $v$。保证每个约数被选择的概率相同。</p>
<p>他想知道 $k$ 次操作后黑板上的数字的期望值是多少。答案对 $10^9+7$ 取模。</p>
<p>数据范围：$1\le n\le 10^{15}$，$1\le k\le 10^4$</p>
</summary>
<category term="数论" scheme="http://hydingsy.github.io/tags/%E6%95%B0%E8%AE%BA/"/>
<category term="动态规划" scheme="http://hydingsy.github.io/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="概率期望" scheme="http://hydingsy.github.io/tags/%E6%A6%82%E7%8E%87%E6%9C%9F%E6%9C%9B/"/>
<category term="积性函数" scheme="http://hydingsy.github.io/tags/%E7%A7%AF%E6%80%A7%E5%87%BD%E6%95%B0/"/>
</entry>
<entry>
<title>「Codeforces 498C」Array and Operations</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-498C-Array-and-Operations/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-498C-Array-and-Operations/</id>
<published>2019-02-24T16:00:00.000Z</published>
<updated>2019-03-03T02:18:59.666Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/498/problem/C" target="_blank" rel="noopener">Codeforces 498C</a></p></blockquote><p>你有一个长度为 $n$ 的正整数序列 $a_1,a_2,\dots,a_n$ 和 $m$ 对整数 $(i_1,j_1),(i_2,j_2),\dots,(i_m,j_m)$。每一对整数 $(i_k,j_k)$ 都满足 $i_k+j_k$ 的值是奇数，且 $1\le i_k<j_k\le n$。</p><p>每一次操作你可以进行一系列行为：</p><ol><li>在 $m$ 个数对中选择一个 $(i_k,j_k)$ 和一个整数 $v$ 满足 $v>1$，并且 $v$ 整除 $a_{i_k}$ 和 $a_{j_k}$。</li><li>将 $a_{i_k}$ 和 $a_{j_k}$ 都除以 $v$。</li></ol><p>请计算出能够进行的最大操作次数。注意，每一对数可能多次使用。</p><p>数据范围：$2\le n\le 100$，$1\le m\le 100$，$1\le a_i\le 10^9$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>首先注意到一个重要的性质：$i_k+j_k$ 为奇数。换言之，我们可以把原序列按照下标奇偶性<strong>分成两部分</strong>，下标为 $i_k$ 和 $j_k$ 的两个数一定不是同一个部分的。也就是说，如果我们在 $i_k$ 和 $j_k$ 之间连边，那么这是一张<strong>二分图</strong>。</p><p>我们建立出这张二分图，新建一个<strong>源点</strong>和<strong>汇点</strong>，每个点向源点或汇点连一条容量为 $\vert S_{a_i}\vert$ 的边（其中 $S_{i}$ 为 $i$ 的质因子可重集合，如 $12$ 的质因子集合为 $\{2,2,3\}$ ），$a_{i_k}$ 和 $a_{j_k}$ 之间的边的容量为 $\vert S_{a_{i_k}}\cap S_{a_{j_k}}\vert$（质因子集合交集大小）。</p><p>看起来直接跑<strong>最大流</strong>就行了？其实很容易找到反例。因为同一个数的质因子不是独立的，一旦被除去之后就不能被别的数字配对。那么我们可以想到<strong>拆点</strong>。</p><ul><li>对于数字 $x$，我们把他拆成<strong>质因子个数个点</strong>，每个质因子向源点或汇点连一条容量为其<strong>指数</strong>的边。</li><li>对于两个数字 $x,y$，我们定义代表的质因子相同的点性质相同。我们把 $x,y$ 性质相同的点连边，容量为这个质因子在 $x$ 和 $y$ 中指数的<strong>较小值</strong>。</li><li>直接跑最大流即可。</li></ul><p><strong>时间复杂度</strong>：$O(m\sqrt n)$，前面应该有一个较大常数，因为点最多有 $9n$ 个，边最多有 $9m$ 条。</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span 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class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span 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class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><queue></span></span></span><br><span class="line"><span class="keyword">typedef</span> <span class="built_in">std</span>::pair<<span class="keyword">int</span>,<span class="keyword">int</span>> pii;</span><br><span class="line"><span class="keyword">typedef</span> <span class="built_in">std</span>::pair<pii,<span class="keyword">int</span>> ppi;</span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e3</span>+<span class="number">5</span>,M=<span class="number">2e5</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,idx,tot=<span class="number">1</span>,lnk[N],cnr[N],ter[M],nxt[M],val[M],dep[N];</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">vector</span><ppi> a[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">add</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> v,<span class="keyword">int</span> w)</span> </span>{</span><br><span class="line"> ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot,val[tot]=w;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">addedge</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> v,<span class="keyword">int</span> w)</span> </span>{</span><br><span class="line"> add(u,v,w),add(v,u,<span class="number">0</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">bfs</span><span class="params">(<span class="keyword">int</span> s,<span class="keyword">int</span> t)</span> </span>{</span><br><span class="line"> <span class="built_in">memset</span>(dep,<span class="number">0</span>,<span class="keyword">sizeof</span>(dep));</span><br><span class="line"> <span class="built_in">memcpy</span>(cnr,lnk,<span class="keyword">sizeof</span>(lnk));</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">queue</span><<span class="keyword">int</span>> q;</span><br><span class="line"> q.push(s),dep[s]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(!q.empty()) {</span><br><span class="line"> <span class="keyword">int</span> u=q.front(); q.pop();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(val[i]&&!dep[v]) q.push(v),dep[v]=dep[u]+<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> dep[t];</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">dfs</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> t,<span class="keyword">int</span> flow)</span> </span>{</span><br><span class="line"> <span class="keyword">if</span>(u==t) <span class="keyword">return</span> flow;</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> &i=cnr[u];i&&ans<flow;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(val[i]&&dep[v]==dep[u]+<span class="number">1</span>) {</span><br><span class="line"> <span class="keyword">int</span> x=dfs(v,t,<span class="built_in">std</span>::min(val[i],flow-ans));</span><br><span class="line"> <span class="keyword">if</span>(x) val[i]-=x,val[i^<span class="number">1</span>]+=x,ans+=x;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(ans<flow) dep[u]=<span class="number">-1</span>;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">dinic</span><span class="params">(<span class="keyword">int</span> s,<span class="keyword">int</span> t)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(bfs(s,t)) {</span><br><span class="line"> <span class="keyword">int</span> x;</span><br><span class="line"> <span class="keyword">while</span>((x=dfs(s,t,<span class="number">1</span><<<span class="number">30</span>))) ans+=x;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">vector</span><ppi> solve(<span class="keyword">int</span> x) {</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">vector</span><ppi> ans;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i*i<=x;++i) {</span><br><span class="line"> <span class="keyword">if</span>(x%i) <span class="keyword">continue</span>;</span><br><span class="line"> <span class="keyword">int</span> cnt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">while</span>(x%i==<span class="number">0</span>) x/=i,++cnt;</span><br><span class="line"> ans.push_back(ppi(pii(i,cnt),++idx));</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span>(x><span class="number">1</span>) ans.push_back(ppi(pii(x,<span class="number">1</span>),++idx));</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&m);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="keyword">int</span> x;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&x);</span><br><span class="line"> a[i]=solve(x);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">while</span>(m--) {</span><br><span class="line"> <span class="keyword">int</span> u,v;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&u,&v);</span><br><span class="line"> <span class="keyword">if</span>(v&<span class="number">1</span>) <span class="built_in">std</span>::swap(u,v);</span><br><span class="line"> <span class="keyword">for</span>(ppi x:a[u]) <span class="keyword">for</span>(ppi y:a[v]) {</span><br><span class="line"> <span class="keyword">if</span>(x.first.first==y.first.first) {</span><br><span class="line"> addedge(x.second,y.second,<span class="built_in">std</span>::min(x.first.second,y.first.second));</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">int</span> s=<span class="number">0</span>,t=idx+<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;i+=<span class="number">2</span>) {</span><br><span class="line"> <span class="keyword">for</span>(ppi x:a[i]) addedge(s,x.second,x.first.second);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i<=n;i+=<span class="number">2</span>) {</span><br><span class="line"> <span class="keyword">for</span>(ppi x:a[i]) addedge(x.second,t,x.first.second);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">int</span> ans=dinic(s,t);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/498/problem/C" target="_blank" rel="noopener">Codeforces 498C</a></p>
</blockquote>
<p>你有一个长度为 $n$ 的正整数序列 $a_1,a_2,\dots,a_n$ 和 $m$ 对整数 $(i_1,j_1),(i_2,j_2),\dots,(i_m,j_m)$。每一对整数 $(i_k,j_k)$ 都满足 $i_k+j_k$ 的值是奇数，且 $1\le i_k&lt;j_k\le n$。</p>
<p>每一次操作你可以进行一系列行为：</p>
<ol>
<li>在 $m$ 个数对中选择一个 $(i_k,j_k)$ 和一个整数 $v$ 满足 $v&gt;1$，并且 $v$ 整除 $a_{i_k}$ 和 $a_{j_k}$。</li>
<li>将 $a_{i_k}$ 和 $a_{j_k}$ 都除以 $v$。</li>
</ol>
<p>请计算出能够进行的最大操作次数。注意，每一对数可能多次使用。</p>
<p>数据范围：$2\le n\le 100$，$1\le m\le 100$，$1\le a_i\le 10^9$</p>
</summary>
<category term="图论" scheme="http://hydingsy.github.io/tags/%E5%9B%BE%E8%AE%BA/"/>
<category term="网络流" scheme="http://hydingsy.github.io/tags/%E7%BD%91%E7%BB%9C%E6%B5%81/"/>
<category term="最大流" scheme="http://hydingsy.github.io/tags/%E6%9C%80%E5%A4%A7%E6%B5%81/"/>
<category term="贪心" scheme="http://hydingsy.github.io/tags/%E8%B4%AA%E5%BF%83/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="二分图" scheme="http://hydingsy.github.io/tags/%E4%BA%8C%E5%88%86%E5%9B%BE/"/>
</entry>
<entry>
<title>「Ynoi 2015」我回来了</title>
<link href="http://hydingsy.github.io/articles/problem-Ynoi-2015-I-am-Back/"/>
<id>http://hydingsy.github.io/articles/problem-Ynoi-2015-I-am-Back/</id>
<published>2019-02-24T16:00:00.000Z</published>
<updated>2019-03-02T06:05:58.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://www.luogu.org/problemnew/show/P5068" target="_blank" rel="noopener">Luogu 5068</a></p></blockquote><p>珂朵莉给你一个 $n$ 个点，$m$ 条边的无向无权图，一共有 $q$ 次询问，每次询问的时候给一堆二元组 $(x_i,y_i)$ 求图中有多少个点 $u$ 与这次询问中至少一个二元组 $(x_i,y_i)$ 满足 $\text{dist}(u,x_i)\le y_i$，$\text{dist}$ 表示这两个点在图中的距离。如果不连通 $\text{dist}=\infin$。</p><p>数据范围：$1\le n\le 10^3$，$1\le m,q\le 10^5$，记二元组的个数和为 $a$ 则 $1\le a\le 2.1\times 10^6$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>我们对每个点 $i$ 和距离 $j$ 维护一个 $\text{bitset}\ f_{i,j}$ 表示和点 $i$ 距离不超过 $j$ 的点的集合。这个不超过可以用 $\text{bitset}$ 的<strong>前缀或</strong>得到。</p><p>询问时只需要把 $f_{x_i,y_i}$ 这些 $\text{bitset}$ 或起来，点集的大小就是答案。</p><p><strong>时间复杂度</strong>：$O(nm+\frac{n^3+an}{\omega})$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><vector></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><bitset></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><queue></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">1e3</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,q,dis[N];</span><br><span class="line"><span class="keyword">bool</span> vis[N];</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">vector</span><<span class="keyword">int</span>> e[N];</span><br><span class="line"><span class="built_in">std</span>::<span class="built_in">bitset</span><N> f[N][N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">bfs</span><span class="params">(<span class="keyword">int</span> s)</span> </span>{</span><br><span class="line"> <span class="built_in">memset</span>(dis,<span class="number">0x3f</span>,<span class="keyword">sizeof</span>(dis));</span><br><span class="line"> <span class="built_in">memset</span>(vis,<span class="number">0</span>,<span class="keyword">sizeof</span>(vis));</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">queue</span><<span class="keyword">int</span>> q;</span><br><span class="line"> q.push(s),dis[s]=<span class="number">0</span>,vis[s]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">while</span>(!q.empty()) {</span><br><span class="line"> <span class="keyword">int</span> u=q.front(); q.pop();</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> v:e[u]) {</span><br><span class="line"> <span class="keyword">if</span>(!vis[v]) dis[v]=dis[u]+<span class="number">1</span>,vis[v]=<span class="number">1</span>,q.push(v);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">init</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> bfs(i);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) {</span><br><span class="line"> <span class="keyword">if</span>(dis[j]!=<span class="number">0x3f3f3f3f</span>) f[i][dis[j]][j]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) f[i][j]|=f[i][j<span class="number">-1</span>];</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d"</span>,&n,&m,&q);</span><br><span class="line"> <span class="keyword">while</span>(m--) {</span><br><span class="line"> <span class="keyword">int</span> u,v;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&u,&v);</span><br><span class="line"> e[u].push_back(v),e[v].push_back(u);</span><br><span class="line"> }</span><br><span class="line"> init();</span><br><span class="line"> <span class="keyword">while</span>(q--) {</span><br><span class="line"> <span class="keyword">int</span> p;</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">bitset</span><N> ans;</span><br><span class="line"> <span class="keyword">for</span>(<span class="built_in">scanf</span>(<span class="string">"%d"</span>,&p);p--;) {</span><br><span class="line"> <span class="keyword">int</span> x,y;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&x,&y);</span><br><span class="line"> ans|=f[x][y>n?n:y];</span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,(<span class="keyword">int</span>)ans.count());</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://www.luogu.org/problemnew/show/P5068" target="_blank" rel="noopener">Luogu 5068</a></p>
</blockquote>
<p>珂朵莉给你一个 $n$ 个点，$m$ 条边的无向无权图，一共有 $q$ 次询问，每次询问的时候给一堆二元组 $(x_i,y_i)$ 求图中有多少个点 $u$ 与这次询问中至少一个二元组 $(x_i,y_i)$ 满足 $\text{dist}(u,x_i)\le y_i$，$\text{dist}$ 表示这两个点在图中的距离。如果不连通 $\text{dist}=\infin$。</p>
<p>数据范围：$1\le n\le 10^3$，$1\le m,q\le 10^5$，记二元组的个数和为 $a$ 则 $1\le a\le 2.1\times 10^6$</p>
</summary>
<category term="图论" scheme="http://hydingsy.github.io/tags/%E5%9B%BE%E8%AE%BA/"/>
<category term="bitset" scheme="http://hydingsy.github.io/tags/bitset/"/>
<category term="Luogu" scheme="http://hydingsy.github.io/tags/Luogu/"/>
<category term="BFS" scheme="http://hydingsy.github.io/tags/BFS/"/>
<category term="最短路" scheme="http://hydingsy.github.io/tags/%E6%9C%80%E7%9F%AD%E8%B7%AF/"/>
</entry>
<entry>
<title>「Codeforces 593E」Strange Calculation and Cats</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-593E-Strange-Calculation-and-Cats/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-593E-Strange-Calculation-and-Cats/</id>
<published>2019-02-23T16:00:00.000Z</published>
<updated>2019-02-28T08:04:02.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/593/problem/E" target="_blank" rel="noopener">Codeforces 593E</a></p></blockquote><p>Gosha 的宇宙是一个由 $n$ 行 $m$ 列组成的表格。他经常被邀请去某个地方，每次接到邀请后，他会先计算到达那个地方的路径方案数，然后再动身出发。Gosha 的房子位于 $(1,1)$。</p><p>在任何时刻，Gosha 都会从当前格子到达相邻的格子，当然不会超出边界。此外，他也可以选择不移动而停留在当前格子里。</p><p>他除了喜欢奇怪的计算，还对猫过敏，因此他从来不去有猫的格子。Gosha 清楚地知道他何时何地会被邀请，也知道猫的在格子上的行程。形式化地讲，他有 $q$ 条记录，第 $i$ 条记录为如下形式之一：</p><ul><li><code>1 x y t</code>：Gosha 在时刻 $t$ 被邀请去格子 $(x,y)$。保证在当前时刻 $(x,y)$ 是没有猫的。</li><li><code>2 x y z</code>：有一只猫会在时刻 $t$ 来到格子 $(x,y)$。保证在当前时刻 $(x,y)$ 是没有别的猫的。</li><li><code>3 x y z</code>：有一只猫会在时刻 $t$ 离开格子 $(x,y)$。保证在当前时刻 $(x,y)$ 是有猫的。</li></ul><p>Gosha 需要你计算出对于每个邀请 $i$，他在时刻 $t_i$ 到达格子 $(x_i,y_i)$ 的方案数。假设他在时刻 $1$ 时从格子 $(1,1)$ 开始移动。</p><p>注意：Gosha 在移动过程中，可以多次访问同一个点，包括 $(1,1)$ 和 $(x_i,y_i)$。由于方案数过大，请将它对 $10^9+7$ 取模。</p><p>数据范围：$1\le n\cdot m\le 20$，$1\le q\le 10^4$，$1\le x_i\le n$，$1\le y_i\le m$，$2\le t_i<t_{i+1}\le 10^9$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>我们把这个 $n\times m$ 的矩阵压缩成一维的，定义 $f_i$ 表示走到 $i$ 格子的方案数。对于当前情况构造转移矩阵 $A$。$A_{i,j}$ 的值为 $1$ 当且仅当可以从 $i$ 走到 $j$。由于时间 $t_i$ 是递增的，直接<strong>矩阵快速幂</strong>求出答案即可。</p><p>注意：当某个点的状态改变（操作 $2$ 或 $3$）时，应该把它的 $f$ 值设为 $0$！</p><p><strong>时间复杂度</strong>：$O(nmq+(nm)^3\log t_i)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstring></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">25</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> mod=<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> dx[]={<span class="number">1</span>,<span class="number">-1</span>,<span class="number">0</span>,<span class="number">0</span>,<span class="number">0</span>},dy[]={<span class="number">0</span>,<span class="number">0</span>,<span class="number">1</span>,<span class="number">-1</span>,<span class="number">0</span>};</span><br><span class="line"><span class="keyword">int</span> h,w,n,q;</span><br><span class="line"><span class="keyword">bool</span> ok[N][N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">upd</span><span class="params">(<span class="keyword">int</span> &x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> (x+=y)>=mod&&(x-=mod);</span><br><span class="line">}</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Matrix</span> {</span></span><br><span class="line"> <span class="keyword">int</span> n,m,A[N][N];</span><br><span class="line"> Matrix(<span class="keyword">int</span> _n=<span class="number">0</span>,<span class="keyword">int</span> _m=<span class="number">0</span>) {</span><br><span class="line"> n=_n,m=_m,<span class="built_in">memset</span>(A,<span class="number">0</span>,<span class="keyword">sizeof</span>(A));</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">void</span> <span class="keyword">operator</span> ~ () {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) A[i][i]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> Matrix <span class="keyword">operator</span> * (<span class="keyword">const</span> Matrix &b)<span class="keyword">const</span> {</span><br><span class="line"> <span class="function">Matrix <span class="title">ans</span><span class="params">(n,b.m)</span></span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=b.m;++j) <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">1</span>;k<=m;++k) {</span><br><span class="line"> upd(ans.A[i][j],<span class="number">1L</span>L*A[i][k]*b.A[k][j]%mod);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line"> }</span><br><span class="line"> Matrix <span class="keyword">operator</span> ^ (<span class="keyword">const</span> <span class="keyword">int</span> &b)<span class="keyword">const</span> {</span><br><span class="line"> Matrix ans(n,n),x=*this; ~ans;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> p=b;p;p>>=<span class="number">1</span>,x=x*x) <span class="keyword">if</span>(p&<span class="number">1</span>) ans=ans*x;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line"> }</span><br><span class="line">};</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">build</span><span class="params">(Matrix &a)</span> </span>{</span><br><span class="line"> <span class="built_in">memset</span>(a.A,<span class="number">0</span>,<span class="keyword">sizeof</span>(a.A));</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=h;++i) <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=w;++j) <span class="keyword">if</span>(ok[i][j]) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">0</span>;k<=<span class="number">4</span>;++k) {</span><br><span class="line"> <span class="keyword">int</span> x=i+dx[k],y=j+dy[k];</span><br><span class="line"> <span class="keyword">if</span>(x>=<span class="number">1</span>&&x<=h&&y>=<span class="number">1</span>&&y<=w&&ok[x][y]) a.A[(i<span class="number">-1</span>)*w+j][(x<span class="number">-1</span>)*w+y]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d"</span>,&h,&w,&q),n=h*w;</span><br><span class="line"> <span class="keyword">int</span> now=<span class="number">1</span>;</span><br><span class="line"> Matrix ans(n,1),a(n,n);</span><br><span class="line"> ans.A[<span class="number">1</span>][<span class="number">1</span>]=<span class="number">1</span>;</span><br><span class="line"> <span class="built_in">memset</span>(ok,<span class="number">1</span>,<span class="keyword">sizeof</span>(ok));</span><br><span class="line"> <span class="keyword">while</span>(q--) {</span><br><span class="line"> <span class="keyword">int</span> opt,x,y,t;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d%d"</span>,&opt,&x,&y,&t);</span><br><span class="line"> build(a),ans=(a^(t-now))*ans,now=t;</span><br><span class="line"> <span class="keyword">if</span>(opt==<span class="number">1</span>) <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans.A[(x<span class="number">-1</span>)*w+y][<span class="number">1</span>]);</span><br><span class="line"> <span class="keyword">if</span>(opt==<span class="number">2</span>) ans.A[(x<span class="number">-1</span>)*w+y][<span class="number">1</span>]=<span class="number">0</span>,ok[x][y]=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">if</span>(opt==<span class="number">3</span>) ans.A[(x<span class="number">-1</span>)*w+y][<span class="number">1</span>]=<span class="number">0</span>,ok[x][y]=<span class="number">1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/593/problem/E" target="_blank" rel="noopener">Codeforces 593E</a></p>
</blockquote>
<p>Gosha 的宇宙是一个由 $n$ 行 $m$ 列组成的表格。他经常被邀请去某个地方，每次接到邀请后，他会先计算到达那个地方的路径方案数，然后再动身出发。Gosha 的房子位于 $(1,1)$。</p>
<p>在任何时刻，Gosha 都会从当前格子到达相邻的格子，当然不会超出边界。此外，他也可以选择不移动而停留在当前格子里。</p>
<p>他除了喜欢奇怪的计算，还对猫过敏，因此他从来不去有猫的格子。Gosha 清楚地知道他何时何地会被邀请，也知道猫的在格子上的行程。形式化地讲，他有 $q$ 条记录，第 $i$ 条记录为如下形式之一：</p>
<ul>
<li><code>1 x y t</code>：Gosha 在时刻 $t$ 被邀请去格子 $(x,y)$。保证在当前时刻 $(x,y)$ 是没有猫的。</li>
<li><code>2 x y z</code>：有一只猫会在时刻 $t$ 来到格子 $(x,y)$。保证在当前时刻 $(x,y)$ 是没有别的猫的。</li>
<li><code>3 x y z</code>：有一只猫会在时刻 $t$ 离开格子 $(x,y)$。保证在当前时刻 $(x,y)$ 是有猫的。</li>
</ul>
<p>Gosha 需要你计算出对于每个邀请 $i$，他在时刻 $t_i$ 到达格子 $(x_i,y_i)$ 的方案数。假设他在时刻 $1$ 时从格子 $(1,1)$ 开始移动。</p>
<p>注意：Gosha 在移动过程中，可以多次访问同一个点，包括 $(1,1)$ 和 $(x_i,y_i)$。由于方案数过大，请将它对 $10^9+7$ 取模。</p>
<p>数据范围：$1\le n\cdot m\le 20$，$1\le q\le 10^4$，$1\le x_i\le n$，$1\le y_i\le m$，$2\le t_i&lt;t_{i+1}\le 10^9$</p>
</summary>
<category term="动态规划" scheme="http://hydingsy.github.io/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="矩阵快速幂" scheme="http://hydingsy.github.io/tags/%E7%9F%A9%E9%98%B5%E5%BF%AB%E9%80%9F%E5%B9%82/"/>
</entry>
<entry>
<title>「Codeforces 1131F」Asya and Kittens</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-1131F-Asya-and-Kittens/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-1131F-Asya-and-Kittens/</id>
<published>2019-02-23T16:00:00.000Z</published>
<updated>2019-03-03T02:41:45.411Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/1131/problem/F" target="_blank" rel="noopener">Codeforces 1131F</a></p></blockquote><p>Asya 最近买了 $n$ 只小猫，把它们标号为 $1$ 到 $n$ 并放在笼子里。笼子由一行 $n$ 个格子组成，从左到右标号为 $1$ 到 $n$。相邻的格子之间透明的隔板，所以起初有 $n-1$ 个隔板。最初每个格子里只有一个只小猫。</p><p>经过观察，Asya 注意到相邻格子中的小猫想要一起玩。因此 Asya 开始删除相邻格子隔板。具体来说，在第 $i$ 天：</p><ul><li>注意到相邻的小猫 $x_i$ 和 $y_i$ 想到一起玩。</li><li>将这两个格子之间的隔板移除，创建了一个新的格子，原来两个格子里的小猫都到这个新的格子里。</li></ul><p>Asya 不会把隔板重新放回去，因此 $n-1$ 天后笼子里只有一个格子，所有的小猫都在里面。</p><p>每一天，Asya 都记得有哪一对小猫 $x_i,y_i$ 想到一起玩，但是她不记得一开始是怎么把小猫放到笼子里的。请帮助她找到小猫在 $n$ 个格子里的可能初始排列。数据保证有解，输出任意一个合法解即可。</p><p>数据范围：$2\le n\le 1.5\times 10^5$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>对于新建一个格子，我们将其看做是<strong>新建了一个点</strong>，这个点的左右儿子为<strong>原来的两个点</strong>（这两个点可能不是叶子，可能是之前合并得到的）。</p><p>这样我们得到了一棵<strong>二叉树</strong>。我们令最后一个点为根进行前序遍历，叶子节点在这个遍历中的序列就是答案。</p><p><strong>时间复杂度</strong>：$O(n\cdot\alpha(n))$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">3e5</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,fa[N],ls[N],rs[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">find</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> fa[x]==x?x:fa[x]=find(fa[x]);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">dfs</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="keyword">if</span>(x<=n) <span class="built_in">printf</span>(<span class="string">"%d "</span>,x);</span><br><span class="line"> <span class="keyword">else</span> dfs(ls[x]),dfs(rs[x]);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&n);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n+n;++i) fa[i]=i;</span><br><span class="line"> <span class="keyword">int</span> m=n;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<n;++i) {</span><br><span class="line"> <span class="keyword">int</span> x,y;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&x,&y);</span><br><span class="line"> x=find(x),y=find(y);</span><br><span class="line"> ls[++m]=x,rs[m]=y;</span><br><span class="line"> fa[x]=fa[y]=m;</span><br><span class="line"> }</span><br><span class="line"> dfs(m);</span><br><span class="line"> <span class="built_in">puts</span>(<span class="string">""</span>);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/1131/problem/F" target="_blank" rel="noopener">Codeforces 1131F</a></p>
</blockquote>
<p>Asya 最近买了 $n$ 只小猫，把它们标号为 $1$ 到 $n$ 并放在笼子里。笼子由一行 $n$ 个格子组成，从左到右标号为 $1$ 到 $n$。相邻的格子之间透明的隔板，所以起初有 $n-1$ 个隔板。最初每个格子里只有一个只小猫。</p>
<p>经过观察，Asya 注意到相邻格子中的小猫想要一起玩。因此 Asya 开始删除相邻格子隔板。具体来说，在第 $i$ 天：</p>
<ul>
<li>注意到相邻的小猫 $x_i$ 和 $y_i$ 想到一起玩。</li>
<li>将这两个格子之间的隔板移除，创建了一个新的格子，原来两个格子里的小猫都到这个新的格子里。</li>
</ul>
<p>Asya 不会把隔板重新放回去，因此 $n-1$ 天后笼子里只有一个格子，所有的小猫都在里面。</p>
<p>每一天，Asya 都记得有哪一对小猫 $x_i,y_i$ 想到一起玩，但是她不记得一开始是怎么把小猫放到笼子里的。请帮助她找到小猫在 $n$ 个格子里的可能初始排列。数据保证有解，输出任意一个合法解即可。</p>
<p>数据范围：$2\le n\le 1.5\times 10^5$</p>
</summary>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="并查集" scheme="http://hydingsy.github.io/tags/%E5%B9%B6%E6%9F%A5%E9%9B%86/"/>
</entry>
<entry>
<title>「Luogu 5171」Earthquake</title>
<link href="http://hydingsy.github.io/articles/problem-Luogu-5171-Earthquake/"/>
<id>http://hydingsy.github.io/articles/problem-Luogu-5171-Earthquake/</id>
<published>2019-02-23T16:00:00.000Z</published>
<updated>2019-03-01T06:54:42.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://www.luogu.org/problemnew/show/P5171" target="_blank" rel="noopener">Luogu 5171</a></p></blockquote><p>给定 $a,b,c$，求满足方程 $ax+by\le c$ 的非负整数解个数。</p><p>数据范围：$1\le a,b\le 10^9$，$0\le c\le \min(a,b)\times 10^9$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>通过<del>简单的</del><strong>线性规划</strong>知识，我们可以得到方程组：</p><script type="math/tex; mode=display">\begin{cases}x\ge 0 \\y\ge 0 \\ax+by\le c\end{cases}</script><p>分析可以得到答案为直线 $ax+by=c$（斜率显然是负数）与 $x$ 轴、$y$ 轴围成的图形中的整点数（包括边界）。也就是答案为（前一部分为 $x$ 轴上的点的个数）：</p><script type="math/tex; mode=display">\left\lfloor\frac{c}{a}\right\rfloor+1+\sum_{i=0}^{\left\lfloor\frac{c}{a}\right\rfloor} \left\lfloor\frac{-ai+c}{b}\right\rfloor</script><p>是不是很像<strong>类欧几里德算法</strong>？具体推导过程详见：<a href="https://orzsiyuan.com/articles/algorithm-Similar-Euclidean-Algorithm/" target="_blank" rel="noopener">「算法笔记」类欧几里德算法</a></p><p>但是这样 $\sum$ 里是会有<strong>负数</strong>的。不信你按照类欧的公式推一下？$f(a,b,c,n)$ 是会递归到 $f\left(c,c-b-1,a,\left\lfloor\frac{an+b}{c}\right\rfloor-1\right)$ 的，这样递归之后直线在 $y$ 轴的截距就是负数了（如果是负数还要和 $0$ 取 $\max$……）。</p><p>我们把这个函数变成增函数。把 $x=0$ 和 $x=\left\lfloor\frac{c}{a}\right\rfloor$ 时函数上的两点对换，得到函数 $y=\frac{ax+c\bmod a}{b}$，故原式化为：</p><script type="math/tex; mode=display">\left\lfloor\frac{c}{a}\right\rfloor+1+\sum_{i=0}^{\left\lfloor\frac{c}{a}\right\rfloor} \left\lfloor\frac{ai+c\bmod a}{b}\right\rfloor</script><p>后半部分就是类欧几里德算法中的函数：</p><script type="math/tex; mode=display">f(a,c\bmod a,b,\left\lfloor\frac{c}{a}\right\rfloor)</script><p>时间复杂度：$O(\log a)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="keyword">typedef</span> <span class="keyword">long</span> <span class="keyword">long</span> LL;</span><br><span class="line"></span><br><span class="line"><span class="function">LL <span class="title">F</span><span class="params">(LL a,LL b,LL c,LL n)</span> </span>{</span><br><span class="line"> <span class="keyword">if</span>(a==<span class="number">0</span>) <span class="keyword">return</span> b/c*(n+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">if</span>(a>=c||b>=c) <span class="keyword">return</span> a/c*n*(n+<span class="number">1</span>)/<span class="number">2</span>+b/c*(n+<span class="number">1</span>)+F(a%c,b%c,c,n);</span><br><span class="line"> LL m=(a*n+b)/c;</span><br><span class="line"> <span class="keyword">return</span> n*m-F(c,c-b<span class="number">-1</span>,a,m<span class="number">-1</span>);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> LL a,b,c;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld%lld%lld"</span>,&a,&b,&c);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%lld\n"</span>,F(a,c%a,b,c/a)+c/a+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://www.luogu.org/problemnew/show/P5171" target="_blank" rel="noopener">Luogu 5171</a></p>
</blockquote>
<p>给定 $a,b,c$，求满足方程 $ax+by\le c$ 的非负整数解个数。</p>
<p>数据范围：$1\le a,b\le 10^9$，$0\le c\le \min(a,b)\times 10^9$</p>
</summary>
<category term="类欧几里德算法" scheme="http://hydingsy.github.io/tags/%E7%B1%BB%E6%AC%A7%E5%87%A0%E9%87%8C%E5%BE%B7%E7%AE%97%E6%B3%95/"/>
<category term="Luogu" scheme="http://hydingsy.github.io/tags/Luogu/"/>
<category term="线性规划" scheme="http://hydingsy.github.io/tags/%E7%BA%BF%E6%80%A7%E8%A7%84%E5%88%92/"/>
</entry>
<entry>
<title>「Codeforces 788B」Weird Journey</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-788B-Weird-Journey/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-788B-Weird-Journey/</id>
<published>2019-02-22T16:00:00.000Z</published>
<updated>2019-02-28T08:28:22.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/788/problem/B" target="_blank" rel="noopener">Codeforces 788B</a></p></blockquote><p>众所周知，Uzhlyandia 有 $n$ 个城市由 $m$ 条道路相连（保证没有重边，但是可能有自环）。定义一条路径是好的，当且仅当这条路径经过 $m-2$ 条边恰好 $2$ 次，另外 $2$ 条边经过恰好 $1$ 次。路径的起始城市和结束城市可以是任意的。</p><p>现在请你计算出有多少条路径是好的，两条路径是不同的当且仅当有一条边不同。</p><p>数据范围：$1\le n,m\le 10^6$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>路径经过 $2$ 次很难处理，我们把每条路径<strong>复制一遍</strong>，这样一来得到：其中 $2$ 条边不经过，其余边恰好经过 $1$ 次。</p><p>换言之，我们把这张新图上的 $2$ 条边删去，这张图存在<strong>欧拉路径</strong>（不多于 $2$ 个点的度数为奇数）。</p><p>新图中，显然每个点的度数都是<strong>偶数</strong>了，我们那么有如下 $3$ 种删边方法：</p><ul><li>删去两个自环（这个点的度数 $-4$）。</li><li>删去一个自环和任意一条普通的边（一个点的度数 $-2$，另外两个点度数各 $-1$）。</li><li>删去两条有共同端点的边（一个点度数 $-2$，另外两个点度数各 $-1$）。</li></ul><p>统计自环个数、边数、度数即可得到答案。</p><p>最后注意一下无解情况！不一定是图不连通，比如 $\text{Codeforces}$ 上的这组数据（应该是有解的）：</p><figure class="highlight basic"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"><span class="symbol">4 </span><span class="number">4</span></span><br><span class="line"><span class="symbol">2 </span><span class="number">3</span></span><br><span class="line"><span class="symbol">2 </span><span class="number">4</span></span><br><span class="line"><span class="symbol">3 </span><span class="number">4</span></span><br><span class="line"><span class="symbol">4 </span><span class="number">4</span></span><br></pre></td></tr></table></figure><p>因此无解当且仅当有连边的点之间相互连通。</p><p><strong>时间复杂度</strong>：$O(m+n\cdot\alpha(n))$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">1e6</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,deg[N],fa[N];</span><br><span class="line"><span class="keyword">bool</span> vis[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">find</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="keyword">return</span> fa[x]==x?x:fa[x]=find(fa[x]);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&m);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) fa[i]=i;</span><br><span class="line"> <span class="keyword">int</span> cnt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=m;++i) {</span><br><span class="line"> <span class="keyword">int</span> u,v;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&u,&v);</span><br><span class="line"> vis[u]=vis[v]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">if</span>(u==v) ++cnt;</span><br><span class="line"> <span class="keyword">else</span> ++deg[u],++deg[v];</span><br><span class="line"> fa[find(u)]=find(v);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">int</span> rt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(vis[i]) {rt=i;<span class="keyword">break</span>;}</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(vis[i]&&find(i)!=find(rt)) <span class="keyword">return</span> <span class="built_in">puts</span>(<span class="string">"0"</span>),<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> ans=<span class="number">0</span>;</span><br><span class="line"> ans+=<span class="number">1L</span>L*cnt*(m-cnt);</span><br><span class="line"> ans+=<span class="number">1L</span>L*cnt*(cnt<span class="number">-1</span>)/<span class="number">2</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) ans+=<span class="number">1L</span>L*deg[i]*(deg[i]<span class="number">-1</span>)/<span class="number">2</span>;</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%lld\n"</span>,ans);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/788/problem/B" target="_blank" rel="noopener">Codeforces 788B</a></p>
</blockquote>
<p>众所周知，Uzhlyandia 有 $n$ 个城市由 $m$ 条道路相连（保证没有重边，但是可能有自环）。定义一条路径是好的，当且仅当这条路径经过 $m-2$ 条边恰好 $2$ 次，另外 $2$ 条边经过恰好 $1$ 次。路径的起始城市和结束城市可以是任意的。</p>
<p>现在请你计算出有多少条路径是好的，两条路径是不同的当且仅当有一条边不同。</p>
<p>数据范围：$1\le n,m\le 10^6$</p>
</summary>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="思维" scheme="http://hydingsy.github.io/tags/%E6%80%9D%E7%BB%B4/"/>
<category term="欧拉路径" scheme="http://hydingsy.github.io/tags/%E6%AC%A7%E6%8B%89%E8%B7%AF%E5%BE%84/"/>
</entry>
<entry>
<title>「IOI 2011」Race</title>
<link href="http://hydingsy.github.io/articles/problem-IOI-2011-Race/"/>
<id>http://hydingsy.github.io/articles/problem-IOI-2011-Race/</id>
<published>2019-02-21T16:00:00.000Z</published>
<updated>2019-03-01T05:03:14.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://www.luogu.org/problemnew/show/P4149" target="_blank" rel="noopener">Luogu 4149</a></p></blockquote><p>给一棵有 $n$ 个节点的树，每条边有权。求一条简单路径，权值和等于 $k$，且边的数量最小。无解输出 $-1$。</p><p>数据范围：$1\le n\le 2\times 10^5$，$1\le k\le 10^6$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p><del>IOI 点分治入门题</del></p><p>为了方便计数，我们采用分别处理每棵子树的方法，而不是容斥（容斥无限 $\text{WA}$ 根本调不出来 QAQ）。</p><p>记录 $f_i$ 表示长度为 $i$ 的链包含的最小边数，对于长度大于 $k$ 的链可以直接忽略。</p><p><strong>时间复杂度</strong>：$O(n\log n)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e5</span>+<span class="number">5</span>,M=<span class="number">4e5</span>+<span class="number">5</span>,K=<span class="number">1e6</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> INF=<span class="number">1</span><<<span class="number">30</span>;</span><br><span class="line"><span class="keyword">int</span> n,k,tot,tp,root,sum,lnk[N],ter[M],nxt[M],val[M],mx[N],sz[N],q[N],f[K],ans;</span><br><span class="line"><span class="keyword">bool</span> vis[N];</span><br><span class="line"><span class="built_in">std</span>::pair<<span class="keyword">int</span>,<span class="keyword">int</span>> s[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">add</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> v,<span class="keyword">int</span> w)</span> </span>{</span><br><span class="line"> ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot,val[tot]=w;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">getRoot</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> fa)</span> </span>{</span><br><span class="line"> sz[u]=<span class="number">1</span>,mx[u]=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(v==fa||vis[v]) <span class="keyword">continue</span>;</span><br><span class="line"> getRoot(v,u);</span><br><span class="line"> sz[u]+=sz[v];</span><br><span class="line"> mx[u]=<span class="built_in">std</span>::max(mx[u],sz[v]);</span><br><span class="line"> }</span><br><span class="line"> mx[u]=<span class="built_in">std</span>::max(mx[u],sum-sz[u]);</span><br><span class="line"> <span class="keyword">if</span>(mx[root]>mx[u]) root=u;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">getDis</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> fa,<span class="keyword">int</span> d,<span class="keyword">int</span> p)</span> </span>{</span><br><span class="line"> <span class="keyword">if</span>(d>k) <span class="keyword">return</span>;</span><br><span class="line"> s[++tp]=<span class="built_in">std</span>::make_pair(d,p);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(v==fa||vis[v]) <span class="keyword">continue</span>;</span><br><span class="line"> getDis(v,u,d+val[i],p+<span class="number">1</span>);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">solve</span><span class="params">(<span class="keyword">int</span> u)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> cnt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(vis[v]) <span class="keyword">continue</span>;</span><br><span class="line"> tp=<span class="number">0</span>,getDis(v,u,val[i],<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=tp;++i) ans=<span class="built_in">std</span>::min(ans,s[i].second+f[k-s[i].first]);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=tp;++i) f[q[++cnt]=s[i].first]=<span class="built_in">std</span>::min(f[s[i].first],s[i].second);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=cnt;++i) f[q[i]]=INF;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">dfs</span><span class="params">(<span class="keyword">int</span> u)</span> </span>{</span><br><span class="line"> vis[u]=<span class="number">1</span>,f[<span class="number">0</span>]=<span class="number">0</span>,solve(u);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(vis[v]) <span class="keyword">continue</span>;</span><br><span class="line"> root=<span class="number">0</span>,sum=sz[v],getRoot(v,u),dfs(root);</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&k);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<n;++i) {</span><br><span class="line"> <span class="keyword">int</span> u,v,w;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d"</span>,&u,&v,&w),++u,++v;</span><br><span class="line"> add(u,v,w),add(v,u,w);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<=k;++i) f[i]=INF;</span><br><span class="line"> ans=mx[<span class="number">0</span>]=INF;</span><br><span class="line"> root=<span class="number">0</span>,sum=n,getRoot(<span class="number">1</span>,<span class="number">0</span>),dfs(root);</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans==INF?<span class="number">-1</span>:ans);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://www.luogu.org/problemnew/show/P4149" target="_blank" rel="noopener">Luogu 4149</a></p>
</blockquote>
<p>给一棵有 $n$ 个节点的树，每条边有权。求一条简单路径，权值和等于 $k$，且边的数量最小。无解输出 $-1$。</p>
<p>数据范围：$1\le n\le 2\times 10^5$，$1\le k\le 10^6$</p>
</summary>
<category term="Luogu" scheme="http://hydingsy.github.io/tags/Luogu/"/>
<category term="IOI" scheme="http://hydingsy.github.io/tags/IOI/"/>
<category term="点分治" scheme="http://hydingsy.github.io/tags/%E7%82%B9%E5%88%86%E6%B2%BB/"/>
</entry>
<entry>
<title>「Codeforces 980E」The Number Games</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-980E-The-Number-Games/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-980E-The-Number-Games/</id>
<published>2019-02-21T16:00:00.000Z</published>
<updated>2019-03-01T01:17:28.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/980/problem/E" target="_blank" rel="noopener">Codeforces 980E</a></p></blockquote><p>Panel 国将举办名为数字游戏的年度表演。每个省派出一名选手。</p><p>国家有 $n$ 个编号从 $1$ 到 $n$ 的省，每个省刚好有一条路径将其与其他省相连。第 $i$ 个省出来的代表有 $2^i$ 名粉丝。</p><p>今年，主席打算削减开支，他想要踢掉 $k$ 个选手。但是，被踢掉的选手的省将很生气，并且不会让别的任何人从这个省经过。</p><p>主席想确保所有剩下选手的省都互相可达，他也希望最大化参与表演的选手的粉丝数。</p><p>主席该踢掉哪些选手呢？</p><p>数据范围：$1\le k<n\le 10^6$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>显然这是一棵树，我们要踢掉的一定是当前度数为 $1$ 的点。换言之，保留的点一定是原树的一个<strong>连通块</strong>。</p><p>再观察 $2^i$ 的性质，我们发现保留一个点 $i$ 一定比保留所有的 $1,2,\dots,i-1$ 更优。那么肯定要尽量删除标号小的点。但是反例如下：</p><blockquote><p>$n=5$，$k=3$，这是一条链：$4-1-5-2-3$</p><p>如果直接删除标号小的，那么删除的点为 $3,2,4$。而最优解为删除 $4,1,3$。</p></blockquote><p>这种贪心过程由于没法考虑到非叶子节点的大小，因此是错误的。那么我们考虑一个反面：尽量<strong>保留标号大的节点</strong>。为了方便操作，我们以 $n$ 作为根节点（因为它一定是要保留的）。具体操作过程如下：</p><ol><li>从大到小遍历节点，假设当前节点为 $x$。我们把已经保留的点叫做<strong>标记点</strong>。</li><li>如果节点 $x$ 到根节点经过的标记点有 $m$ 个，那么如果要保留点 $x$，还需要额外保留它到根路径上的非标记点 $dep_x-m-1$ 个（其中 $dep$ 为深度，根的深度为 $1$），共需要保留 $dep_x-m$ 个。</li><li>如果 $dep_x-m$ 不大于当前允许保留的节点个数，那么把这些点全部保留；否则不操作。</li><li>跳到第 $1$ 步。</li></ol><p>最后考虑实现问题：</p><ul><li><p>如何维护标记点个数？</p><p>我们发现每加入一个标记点 $x$，$x$ 子树内的节点到根的标记点个数都会增加 $1$，直接求出 $\text{DFS}$ 序，用<strong>树状数组</strong>维护即可。</p></li><li><p>如何把 $dep_x-m$ 个点设为标记点。</p><p>暴力往上跳即可，跳到标记点就退出。每个点最多经过 $1$ 次，复杂度正确。</p></li></ul><p><strong>时间复杂度</strong>：$O(n\log n)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">1e6</span>+<span class="number">5</span>,M=<span class="number">2e6</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,k,tot,idx,lnk[N],ter[M],nxt[M],dfn[N],dep[N],fa[N],sz[N],b[N];</span><br><span class="line"><span class="keyword">bool</span> flg[N];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">add</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> v)</span> </span>{</span><br><span class="line"> ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">dfs</span><span class="params">(<span class="keyword">int</span> u,<span class="keyword">int</span> p)</span> </span>{</span><br><span class="line"> dfn[u]=++idx,dep[u]=dep[fa[u]=p]+<span class="number">1</span>,sz[u]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=lnk[u];i;i=nxt[i]) {</span><br><span class="line"> <span class="keyword">int</span> v=ter[i];</span><br><span class="line"> <span class="keyword">if</span>(v==p) <span class="keyword">continue</span>;</span><br><span class="line"> dfs(v,u);</span><br><span class="line"> sz[u]+=sz[v];</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">modify</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> v)</span> </span>{</span><br><span class="line"> <span class="keyword">for</span>(;x<=n;x+=x&-x) b[x]+=v;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">query</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(;x;x^=x&-x) ans+=b[x];</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&k);</span><br><span class="line"> k=n-k;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<n;++i) {</span><br><span class="line"> <span class="keyword">int</span> u,v;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&u,&v);</span><br><span class="line"> add(u,v),add(v,u);</span><br><span class="line"> }</span><br><span class="line"> dfs(n,<span class="number">0</span>);</span><br><span class="line"> flg[n]=<span class="number">1</span>,--k;</span><br><span class="line"> modify(dfn[n],<span class="number">1</span>),modify(dfn[n]+sz[n],<span class="number">-1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=n<span class="number">-1</span>;i>=<span class="number">1</span>;--i) {</span><br><span class="line"> <span class="keyword">if</span>(!flg[i]&&dep[i]-query(dfn[i])<=k) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> x=i;x!=n&&!flg[x];x=fa[x]) {</span><br><span class="line"> modify(dfn[x],<span class="number">1</span>),modify(dfn[x]+sz[x],<span class="number">-1</span>);</span><br><span class="line"> flg[x]=<span class="number">1</span>,--k;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="keyword">if</span>(!flg[i]) <span class="built_in">printf</span>(<span class="string">"%d "</span>,i);</span><br><span class="line"> <span class="built_in">puts</span>(<span class="string">""</span>);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/980/problem/E" target="_blank" rel="noopener">Codeforces 980E</a></p>
</blockquote>
<p>Panel 国将举办名为数字游戏的年度表演。每个省派出一名选手。</p>
<p>国家有 $n$ 个编号从 $1$ 到 $n$ 的省，每个省刚好有一条路径将其与其他省相连。第 $i$ 个省出来的代表有 $2^i$ 名粉丝。</p>
<p>今年，主席打算削减开支，他想要踢掉 $k$ 个选手。但是，被踢掉的选手的省将很生气，并且不会让别的任何人从这个省经过。</p>
<p>主席想确保所有剩下选手的省都互相可达，他也希望最大化参与表演的选手的粉丝数。</p>
<p>主席该踢掉哪些选手呢？</p>
<p>数据范围：$1\le k&lt;n\le 10^6$</p>
</summary>
<category term="数据结构" scheme="http://hydingsy.github.io/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84/"/>
<category term="贪心" scheme="http://hydingsy.github.io/tags/%E8%B4%AA%E5%BF%83/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="树状数组" scheme="http://hydingsy.github.io/tags/%E6%A0%91%E7%8A%B6%E6%95%B0%E7%BB%84/"/>
<category term="DFS 序" scheme="http://hydingsy.github.io/tags/DFS-%E5%BA%8F/"/>
</entry>
<entry>
<title>「Codeforces 555B」Case of Fugitive</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-555B-Case-of-Fugitive/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-555B-Case-of-Fugitive/</id>
<published>2019-02-21T16:00:00.000Z</published>
<updated>2019-02-28T07:41:20.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/555/problem/B" target="_blank" rel="noopener">Codeforces 555B</a></p></blockquote><p>有 $n$ 个狭窄的岛屿排成一排的群岛，我们将他们表示为直线上不相交的线段：岛屿 $i$ 的坐标为 $[l_i,r_i]$。保证对于所有的 $1\le i\le n-1$，都有 $r_i<l_{i+1}$。</p><p>你需要在所有相邻的岛屿之间架上桥梁。一座长度为 $a$ 的桥可以架在第 $i$ 和 $i+1$ 个岛屿之间，当且仅当存在 $l_i\le x\le r_i$，$l_{i+1}\le y\le r_{i+1}$ 满足 $y-x=a$。</p><p>现在你有 $m$ 座桥梁，第 $i$ 座桥梁的长度为 $a_i$，并且每座桥梁最多只能使用一次。请问是否可以把任意两个相邻的岛屿连通。如果可行，输出 <code>Yes</code> 和方案；否则输出 <code>No</code>。</p><p>数据范围：$2\le n\le 2\times 10^5$，$1\le m\le 2\times 10^5$，$1\le l_i\le r_i\le 10^{18}$，$1\le a_i\le 10^{18}$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>我们先求出相邻两个岛屿之间能架的桥的长度区间，显然对于岛屿 $i,i+1$，长度区间为 $[l_{i+1}-r_i,r_{i+1}-l_i]$，记为 $[s_i,t_i]$。这就是说，我们要对每个 $i\in [1,n)$，配上一个 $a_j$ 满足 $s_i\le a_j\le t_i$。</p><p>考虑贪心策略：把所有的长度区间按照<strong>右端点</strong>为第一关键字，<strong>左端点</strong>为第二关键字，从小到大排序。对于第 $i$ 个区间，我们找到第一个不小于 $s_i$ 的 $a_j$ 与其配对。如果出现无解情况（找不到不小于 $s_i$ 的值，或者 $a_j>r_i$）就直接退出。这个过程可以通过 $\text{set}$ 实现。</p><p><strong>时间复杂度</strong>：$O(n\log nm)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><set></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">2e5</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,ans[N];</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Inter</span> {</span></span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> l,r;</span><br><span class="line"> <span class="keyword">int</span> id;</span><br><span class="line"> <span class="keyword">bool</span> <span class="keyword">operator</span> < (<span class="keyword">const</span> Inter &b)<span class="keyword">const</span> {</span><br><span class="line"> <span class="keyword">return</span> r==b.r?l<b.l:r<b.r;</span><br><span class="line"> }</span><br><span class="line">} a[N];</span><br><span class="line"><span class="class"><span class="keyword">struct</span> <span class="title">Bridge</span> {</span></span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> x;</span><br><span class="line"> <span class="keyword">int</span> id;</span><br><span class="line"> <span class="keyword">bool</span> <span class="keyword">operator</span> < (<span class="keyword">const</span> Bridge &b)<span class="keyword">const</span> {</span><br><span class="line"> <span class="keyword">return</span> x<b.x;</span><br><span class="line"> }</span><br><span class="line">};</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&n,&m);</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> lstL,lstR;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld%lld"</span>,&lstL,&lstR);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<n;++i) {</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> L,R;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld%lld"</span>,&L,&R);</span><br><span class="line"> a[i].l=L-lstR,a[i].r=R-lstL,a[i].id=i;</span><br><span class="line"> lstL=L,lstR=R;</span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">multiset</span><Bridge> s;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=m;++i) {</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> x;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%lld"</span>,&x);</span><br><span class="line"> s.insert(Bridge{x,i});</span><br><span class="line"> }</span><br><span class="line"> --n;</span><br><span class="line"> <span class="built_in">std</span>::sort(a+<span class="number">1</span>,a+n+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) {</span><br><span class="line"> <span class="keyword">long</span> <span class="keyword">long</span> l=a[i].l,r=a[i].r;</span><br><span class="line"> <span class="built_in">std</span>::<span class="built_in">multiset</span><Bridge>::iterator it=s.lower_bound(Bridge{l,<span class="number">0</span>});</span><br><span class="line"> <span class="keyword">if</span>(it==s.end()||it->x>r) <span class="keyword">return</span> <span class="built_in">puts</span>(<span class="string">"No"</span>),<span class="number">0</span>;</span><br><span class="line"> ans[a[i].id]=it->id,s.erase(it);</span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">puts</span>(<span class="string">"Yes"</span>);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="built_in">printf</span>(<span class="string">"%d%c"</span>,ans[i],<span class="string">" \n"</span>[i==n]);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/555/problem/B" target="_blank" rel="noopener">Codeforces 555B</a></p>
</blockquote>
<p>有 $n$ 个狭窄的岛屿排成一排的群岛，我们将他们表示为直线上不相交的线段：岛屿 $i$ 的坐标为 $[l_i,r_i]$。保证对于所有的 $1\le i\le n-1$，都有 $r_i&lt;l_{i+1}$。</p>
<p>你需要在所有相邻的岛屿之间架上桥梁。一座长度为 $a$ 的桥可以架在第 $i$ 和 $i+1$ 个岛屿之间，当且仅当存在 $l_i\le x\le r_i$，$l_{i+1}\le y\le r_{i+1}$ 满足 $y-x=a$。</p>
<p>现在你有 $m$ 座桥梁，第 $i$ 座桥梁的长度为 $a_i$，并且每座桥梁最多只能使用一次。请问是否可以把任意两个相邻的岛屿连通。如果可行，输出 <code>Yes</code> 和方案；否则输出 <code>No</code>。</p>
<p>数据范围：$2\le n\le 2\times 10^5$，$1\le m\le 2\times 10^5$，$1\le l_i\le r_i\le 10^{18}$，$1\le a_i\le 10^{18}$</p>
</summary>
<category term="贪心" scheme="http://hydingsy.github.io/tags/%E8%B4%AA%E5%BF%83/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
</entry>
<entry>
<title>「Codeforces 1063C」Dwarves, Hats and Extrasensory Abilities</title>
<link href="http://hydingsy.github.io/articles/problem-Codeforces-1063C-Dwarves-Hats-and-Extrasensory-Abilities/"/>
<id>http://hydingsy.github.io/articles/problem-Codeforces-1063C-Dwarves-Hats-and-Extrasensory-Abilities/</id>
<published>2019-02-20T16:00:00.000Z</published>
<updated>2019-02-28T03:23:40.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://codeforces.com/contest/1063/problem/C" target="_blank" rel="noopener">Codeforces 1063C</a></p></blockquote><p>这是一道交互题。</p><p>给你一个正整数 $n$ 表示有 $n$ 个点。接下来你可以询问任意 $n$ 个点 $(x,y)$ 满足 $0\le x,y\le 10^9$，交互器会告诉你每个点的颜色是黑色还是白色。你需要确定一条直线，使得直线将所有的点分成两个部分，一部分的点都是白色，另一部分都是黑色。</p><p>数据范围：$1\le n\le 30$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>由于是任意询问点，我们可以使所有的点<strong>都在一条直线上</strong>，左侧、右侧的点颜色不同。</p><p>首先询问点 $(0,y)$（这里的 $y$ 可以随意确定，只不过是保证点连成的直线和 $x$ 轴平行罢了），假设得到的颜色为白色，那么我们通过某种构造方法，使得左侧的点都是白色，右侧都是黑色。注意：接下来的分析都依次为根据！</p><p>定义搜索区间为<strong>最右侧的白点</strong>到<strong>最左侧的黑点</strong>形成的区间（不包含两端）$[l,r]$。每次询问区间中点 $mid$，如果 $(mid,y)$ 的颜色为白色，那么搜索区间变为 $(mid,r]$ 否则为 $[l,mid)$。</p><p>这样我们最多可以询问 $\left\lfloor\log_2 10^9\right\rfloor=30$ 个点，刚好达到 $n$ 的上限。</p><p>画图可知，通过 $(l,0)$ 和 $(r,2)$ 的直线是满足要求的。（千万不要使用 $x=\left\lfloor\frac{l+r}{2}\right\rfloor$ 这条直线，在 $r-l\le 1$ 时会出错！）</p><p><strong>时间复杂度</strong>：$O(n)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><algorithm></span></span></span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">ask</span><span class="params">(<span class="keyword">int</span> x)</span> </span>{</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d 1\n"</span>,x),fflush(<span class="built_in">stdout</span>);</span><br><span class="line"> <span class="keyword">char</span> s[<span class="number">10</span>];</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%s"</span>,s+<span class="number">1</span>);</span><br><span class="line"> <span class="keyword">return</span> s[<span class="number">1</span>]==<span class="string">'w'</span>;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> n;</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d"</span>,&n);</span><br><span class="line"> <span class="keyword">int</span> col=ask(<span class="number">0</span>);</span><br><span class="line"> <span class="keyword">int</span> l=<span class="number">1</span>,r=<span class="number">1e9</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">2</span>;i<=n;++i) {</span><br><span class="line"> <span class="keyword">int</span> mid=(l+r)>><span class="number">1</span>;</span><br><span class="line"> <span class="keyword">if</span>(ask(mid)==col) l=mid+<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">else</span> r=mid<span class="number">-1</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d 0 %d 2\n"</span>,l,r);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://codeforces.com/contest/1063/problem/C" target="_blank" rel="noopener">Codeforces 1063C</a></p>
</blockquote>
<p>这是一道交互题。</p>
<p>给你一个正整数 $n$ 表示有 $n$ 个点。接下来你可以询问任意 $n$ 个点 $(x,y)$ 满足 $0\le x,y\le 10^9$，交互器会告诉你每个点的颜色是黑色还是白色。你需要确定一条直线，使得直线将所有的点分成两个部分，一部分的点都是白色，另一部分都是黑色。</p>
<p>数据范围：$1\le n\le 30$</p>
</summary>
<category term="交互题" scheme="http://hydingsy.github.io/tags/%E4%BA%A4%E4%BA%92%E9%A2%98/"/>
<category term="Codeforces" scheme="http://hydingsy.github.io/tags/Codeforces/"/>
<category term="二分" scheme="http://hydingsy.github.io/tags/%E4%BA%8C%E5%88%86/"/>
</entry>
<entry>
<title>「Luogu 5196」Cow Poetry</title>
<link href="http://hydingsy.github.io/articles/problem-Luogu-5196-Cow-Poetry/"/>
<id>http://hydingsy.github.io/articles/problem-Luogu-5196-Cow-Poetry/</id>
<published>2019-02-16T16:00:00.000Z</published>
<updated>2019-02-27T06:12:24.000Z</updated>
<content type="html"><![CDATA[<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote><p>题目链接：<a href="https://www.luogu.org/problemnew/show/P5196" target="_blank" rel="noopener">Luogu 5196</a></p></blockquote><p>不为 Farmer John 所知的是，Bessie 还热衷于资助艺术创作！最近，她开始研究许多伟大的诗人们，而现在，她想要尝试创作一些属于自己的诗歌了。Bessie 认识 $n$ 个单词，她想要将她们写进她的诗。Bessie 已经计算了她认识的每个单词的长度 $s_i$，以音节为单位，并且她将这些单词划分成了不同的“韵部”，第 $i$ 个单词的韵部为 $c_i$。每个单词仅与属于同一韵部的其他单词押韵。</p><p>Bessie 的每首诗由 $m$ 行组成，每一行必须由 $k$ 个音节构成。此外，Bessie 的诗必须遵循某个指定的押韵模式。每行有一个押韵模式，用大写字母 $e_i$ 表示。所有押韵模式等于 $e_i$ 的行必须以同一韵部的单词结尾。不同 $e_i$ 值的行并非必须以不同的韵部的单词结尾。</p><p>Bessie 想要知道她可以写出多少首符合限制条件的不同的诗，答案对 $10^9+7$ 取模。</p><p>数据范围：$1\le n,k\le 5\times 10^3$，$1\le m\le 10^5$，$1\le s_i\le k$，$1\le c_i\le n$</p><a id="more"></a><hr><h2 id="Solution"><a href="#Solution" class="headerlink" title="Solution"></a>Solution</h2><p>很显然是一个 $\text{DP}$ 计数题。定义状态 $f(i,j)$ 表示长度为 $i$ 的句子，以 $j$ 韵部结尾的方案数。</p><p>我们枚举当前长度 $i$，以韵部 $j$ 结尾，上一个单词的韵部 $k$，那么有转移：</p><script type="math/tex; mode=display">f(i,j)=\sum_{x=1,c_x=j}^n\sum_{k=1}^n f(i-s_x,k)</script><p>略作优化后写成伪代码就是：</p><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=k;++i) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) {</span><br><span class="line"> <span class="keyword">if</span>(i<s[j]) <span class="keyword">continue</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">1</span>;k<=n;++k) {</span><br><span class="line"> f[i][c[j]]+=f[i-s[j]][k];</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure><p>很显然这个转移是 $O(n^2k)$ 的，无法接受。</p><p>我们发现最外面的 $i$ 和 $j$ 的枚举是没法去掉了，试图把内层的 $k$ 的循环优化掉。设 $g(i)$ 表示长度为 $i$ 的总方案数：</p><script type="math/tex; mode=display">g(i)=\sum_{k=1}^n f(i,k)</script><p>那么我们的转移就是：</p><script type="math/tex; mode=display">f(i,j)=\sum_{x=1,c_x=j}^n g(i-s_x)</script><p>最后考虑句子的统计。记韵部为 $i$ 的句子有 $cnt_i$ 个，那么答案就是：</p><script type="math/tex; mode=display">\prod_{i=1}^{26}\sum_{j=1}^n f(k,j)^{cnt_i}</script><p><strong>时间复杂度</strong>：$O(nk)$</p><hr><h2 id="Code"><a href="#Code" class="headerlink" title="Code"></a>Code</h2><figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cstdio></span></span></span><br><span class="line"></span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> N=<span class="number">5e3</span>+<span class="number">5</span>;</span><br><span class="line"><span class="keyword">const</span> <span class="keyword">int</span> mod=<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"><span class="keyword">int</span> n,m,k,l[N],c[N],f[N][N],g[N],cnt[<span class="number">26</span>];</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">void</span> <span class="title">upd</span><span class="params">(<span class="keyword">int</span> &x,<span class="keyword">int</span> y)</span> </span>{</span><br><span class="line"> (x+=y)>=mod&&(x-=mod);</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">char</span> <span class="title">getc</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="keyword">char</span> c=getchar();</span><br><span class="line"> <span class="keyword">while</span>(c<<span class="string">'A'</span>||c><span class="string">'Z'</span>) c=getchar();</span><br><span class="line"> <span class="keyword">return</span> c;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">pow</span><span class="params">(<span class="keyword">int</span> x,<span class="keyword">int</span> p)</span> </span>{</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(;p;p>>=<span class="number">1</span>,x=<span class="number">1L</span>L*x*x%mod) <span class="keyword">if</span>(p&<span class="number">1</span>) ans=<span class="number">1L</span>L*ans*x%mod;</span><br><span class="line"> <span class="keyword">return</span> ans;</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span> </span>{</span><br><span class="line"> <span class="built_in">scanf</span>(<span class="string">"%d%d%d"</span>,&n,&m,&k);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=n;++i) <span class="built_in">scanf</span>(<span class="string">"%d%d"</span>,&l[i],&c[i]);</span><br><span class="line"> g[<span class="number">0</span>]=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=k;++i) {</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) <span class="keyword">if</span>(i>=l[j]) upd(f[i][c[j]],g[i-l[j]]);</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) upd(g[i],f[i][j]);</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=m;++i) ++cnt[getc()-<span class="string">'A'</span>];</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">1</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<<span class="number">26</span>;++i) <span class="keyword">if</span>(cnt[i]) {</span><br><span class="line"> <span class="keyword">int</span> sum=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">1</span>;j<=n;++j) upd(sum,<span class="built_in">pow</span>(f[k][j],cnt[i]));</span><br><span class="line"> ans=<span class="number">1L</span>L*ans*sum%mod;</span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">printf</span>(<span class="string">"%d\n"</span>,ans);</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>]]></content>
<summary type="html">
<h2 id="Description"><a href="#Description" class="headerlink" title="Description"></a>Description</h2><blockquote>
<p>题目链接：<a href="https://www.luogu.org/problemnew/show/P5196" target="_blank" rel="noopener">Luogu 5196</a></p>
</blockquote>
<p>不为 Farmer John 所知的是，Bessie 还热衷于资助艺术创作！最近，她开始研究许多伟大的诗人们，而现在，她想要尝试创作一些属于自己的诗歌了。Bessie 认识 $n$ 个单词，她想要将她们写进她的诗。Bessie 已经计算了她认识的每个单词的长度 $s_i$，以音节为单位，并且她将这些单词划分成了不同的“韵部”，第 $i$ 个单词的韵部为 $c_i$。每个单词仅与属于同一韵部的其他单词押韵。</p>
<p>Bessie 的每首诗由 $m$ 行组成，每一行必须由 $k$ 个音节构成。此外，Bessie 的诗必须遵循某个指定的押韵模式。每行有一个押韵模式，用大写字母 $e_i$ 表示。所有押韵模式等于 $e_i$ 的行必须以同一韵部的单词结尾。不同 $e_i$ 值的行并非必须以不同的韵部的单词结尾。</p>
<p>Bessie 想要知道她可以写出多少首符合限制条件的不同的诗，答案对 $10^9+7$ 取模。</p>
<p>数据范围：$1\le n,k\le 5\times 10^3$，$1\le m\le 10^5$，$1\le s_i\le k$，$1\le c_i\le n$</p>
</summary>
<category term="动态规划" scheme="http://hydingsy.github.io/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/"/>
<category term="Luogu" scheme="http://hydingsy.github.io/tags/Luogu/"/>
<category term="背包" scheme="http://hydingsy.github.io/tags/%E8%83%8C%E5%8C%85/"/>
</entry>
</feed>